CHEMN--12-N08/4/CHEMI/HP2/ENG/TZO/XX(Question 6 co

Solution

1

Identify the reducing agent: To find a species that can reduce bromine (

\mathrm{Br}_2

) to bromide ions (

\mathrm{Br}^-

) but not iodine (

\mathrm{I}_2

) to iodide ions (

\mathrm{I}^-

), we look for a species with a standard electrode potential that is more positive than that of

\mathrm{Br}_2/\mathrm{Br}^-

but less positive than that of

\mathrm{I}_2/\mathrm{I}^-

. From standard electrode potential tables,

\mathrm{Fe}^{3+}

can reduce

\mathrm{Br}_2

but not

\mathrm{I}_2

. The half-reaction for this reduction is:

\mathrm{Fe}^{3+} + \mathrm{Br}_2 + 2\mathrm{e}^- \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{Br}^-

2

Calculate the cell potential: The cell potential (

E^\circ_{\text{cell}}

) for the cell with aluminium and silver can be calculated using the standard electrode potentials of the half-reactions. The half-reaction for aluminium is

\mathrm{Al}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Al}

and for silver is

\mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag}

. The standard potentials are approximately

-1.66 \, \text{V}

for aluminium and

+0.80 \, \text{V}

for silver. Thus, the cell potential is:

E^\circ_{\text{cell}} = E^\circ_{\text{Ag}} - E^\circ_{\text{Al}} = 0.80 - (-1.66) = 2.46 \, \text{V}.

3

Deduce the standard electrode potential: Given the cell potential of

+0.48 \, \text{V}

for the cell

\mathrm{Zn}(s) | \mathrm{Zn}^{2+}(aq) || \mathrm{Co}^{2+}(aq) | \mathrm{Co}(s)

, we can deduce the standard electrode potential for the cobalt half-reaction. The half-reaction for cobalt is

\mathrm{Co}^{2+} + 2\mathrm{e}^- \rightleftharpoons \mathrm{Co}

. Using the known cell potential, we can express it as:

E^\circ_{\text{cell}} = E^\circ_{\text{Co}} - E^\circ_{\text{Zn}}.

Assuming

E^\circ_{\text{Zn}} \approx -0.76 \, \text{V}

, we can rearrange to find

E^\circ_{\text{Co}}

:

E^\circ_{\text{Co}} = E^\circ_{\text{cell}} + E^\circ_{\text{Zn}} = 0.48 + 0.76 = 1.24 \, \text{V}.

4

Explain water as a ligand: Water acts as a ligand by donating lone pairs of electrons from its oxygen atom to form coordinate bonds with metal ions. In the complexes

\left[\mathrm{Zn}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+}

and

\left[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}

, water molecules surround the metal ions, forming octahedral and tetrahedral geometries, respectively. The shape of

\left[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}

is octahedral

5

Explain color differences: The color of a solution containing

\left[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}

arises from d-d transitions of electrons in the cobalt ion, which is a transition metal with partially filled d-orbitals. In contrast,

\left[\mathrm{Zn}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+}

has a full d-subshell and does not exhibit such transitions, resulting in a colorless solution

Answer

The reducing agent is

\mathrm{Fe}^{3+}

; the cell potential for Al and Ag is

2.46 \, \text{V}

; the standard potential for Co is

1.24 \, \text{V}

; water acts as a ligand forming octahedral complexes;

\left[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}

is colored due to d-d transitions, while

\left[\mathrm{Zn}\left(\mathrm{H}_{2}\mathrm{O}\right)_{4}\right]^{2+}

is colorless.

Key Concept

Standard electrode potentials and ligand behavior in coordination complexes

Explanation

The answers detail the identification of reducing agents, calculation of cell potentials, and the role of water as a ligand, highlighting the differences in color due to electronic transitions in transition metals.

Solution

1

Identify compound A: The molecular formula

C_4H_8O_2

suggests that compound A could be an ester or a carboxylic acid. Given the context, a likely candidate is butanoic acid, which has the structural formula

CH_3(CH_2)_{2}COOH

2

Functional groups in B, C, and D: Compounds B, C, and D likely contain the carboxylic acid functional group (

-COOH

) or an ester group (

-COOR

), as they share the same molecular formula

3

Reactions of butane with bromine: The reaction of butane with bromine involves free radical substitution, where bromine radicals (

Br\cdot

) are generated. The

S_N2

reaction involves a nucleophile attacking the carbon atom, leading to the formation of a new bond

4

Bond fission types: The two types of bond fission are homolytic fission, which produces free radicals, and heterolytic fission, which produces ions

5

Meaning of $S_N2$ symbols: The

S_N2

stands for Substitution Nucleophilic Bimolecular, indicating a two-step process where the nucleophile attacks the substrate simultaneously as the leaving group departs

6

Mechanism of $S_N2$ reaction: The mechanism involves the nucleophile attacking the electrophilic carbon, forming a transition state where both the nucleophile and leaving group are partially bonded to the carbon. Curly arrows can be used to illustrate this movement

7

Intermediate in $S_N1$ reaction: The intermediate formed in the

S_N1

reaction is a carbocation, which can be represented as

C^+

where the carbon has a positive charge

8

Structure of compound G: Compound G can be deduced as a brominated derivative of butane, likely 2-bromobutane, formed by the reaction of butane with bromine. The reagents used would be

Br_2

and a suitable solvent

9

Isomer of compound G: An isomer of compound G that can be converted into compound A is 2-bromobutane, which can undergo elimination to form butanoic acid

10

Boiling point prediction: Compound F (butan-2-ol) likely has the highest boiling point due to hydrogen bonding, which is the strongest intermolecular force present

11

Lowest solubility prediction: Compound E is expected to have the lowest solubility in water due to weaker van der Waals forces compared to hydrogen bonding in butan-2-ol

12

Optical isomers in butan-2-ol: The presence of a chiral center in butan-2-ol leads to optical isomerism. The two isomers can be distinguished using plane polarized light, which rotates the light in different directions depending on the isomer

Answer

Compound A is butanoic acid; compounds B, C, and D contain carboxylic acid or ester groups; the reaction of butane with bromine involves free radicals; the

S_N2

mechanism involves a bimolecular substitution; the intermediate in the

S_N1

reaction is a carbocation; compound G is likely 2-bromobutane; butan-2-ol has the highest boiling point due to hydrogen bonding; compound E has the lowest solubility due to weaker forces; butan-2-ol exhibits optical isomerism.

Key Concept

Understanding organic reactions and properties of compounds based on their structures and functional groups.

Explanation

The answers provide insights into the structural characteristics, reaction mechanisms, and physical properties of organic compounds derived from butane.

Solution

1

Discuss the formation of bonds in ethyne: In ethyne (

\mathrm{C}_2\mathrm{H}_2

), the carbon atoms undergo

sp

hybridization, forming two

sp

hybrid orbitals that create a sigma bond (

\sigma

) between the two carbon atoms. The remaining two unhybridized

p

orbitals on each carbon atom overlap to form two pi bonds (

\pi

), resulting in a triple bond between the carbon atoms. Thus, ethyne has one

\sigma

bond and two

\pi

bonds

2

Identify hybridization in cyclohexene: Cyclohexene (

\mathrm{C}_6\mathrm{H}_{10}

) features

sp^2

hybridization at the double bond carbon atoms, leading to bond angles of approximately

120^\circ

. The remaining carbon atoms are

sp^3

hybridized, with bond angles around

109.5^\circ

3

Explain the representation of benzene: The symbol

\bigcirc

is often used to represent benzene due to its resonance structure, which indicates equal bond lengths between carbon atoms. The data shows that the enthalpy change for hydrogenation of benzene is less than that of cyclohexene, suggesting that benzene is more stable and has delocalized electrons

4

Discuss bond lengths in benzene: The carbon-to-carbon bond lengths in benzene are equal and shorter than typical single bonds, supporting the use of the

\bigcirc

symbol, which represents the delocalization of electrons across the ring structure

5

Discuss bonding in benzoate ion: In the benzoate ion (

\mathrm{C}_6\mathrm{H}_5\mathrm{COO}^-

), the

\mathrm{COO}^-

part exhibits resonance, where the negative charge is delocalized between the two oxygen atoms, leading to equal bond lengths between carbon and both oxygen atoms

6

Compare bond lengths in benzoic acid and benzoate ion: The carbon-to-oxygen bond in benzoic acid (

\mathrm{C}_6\mathrm{H}_5\mathrm{COOH}

) is typically shorter than in the benzoate ion due to the presence of a double bond in benzoic acid, while the resonance in the benzoate ion leads to longer, equivalent bond lengths

7

Explain reaction likelihood: Reaction I (

\mathrm{C}_6\mathrm{H}_6 + \mathrm{Cl}_2 \rightarrow \mathrm{C}_6\mathrm{H}_5\mathrm{Cl} + \mathrm{HCl}

) is more likely to occur than Reaction II (

\mathrm{C}_6\mathrm{H}_6 + \mathrm{Cl}_2 \rightarrow \mathrm{C}_6\mathrm{H}_6\mathrm{Cl}_2

) because the first reaction involves electrophilic substitution, which is favored in aromatic compounds like benzene, while the second reaction would require a more reactive condition

Answer

Ethyne has a triple bond with one sigma and two pi bonds; cyclohexene has

sp^2

and

sp^3

hybridization; benzene is best represented by

\bigcirc

due to resonance; bond lengths support this; benzoate ion shows delocalization; benzoic acid has shorter C=O bonds than benzoate; reaction I is favored due to electrophilic substitution.

Key Concept

Hybridization and resonance structures are crucial in understanding molecular bonding and stability in organic compounds.

Explanation

The answer explains the types of bonds and hybridization in ethyne and cyclohexene, the representation of benzene, and the bonding characteristics in benzoate ion, highlighting the stability and reactivity of these compounds.

Solution

1

Calculate the acid dissociation constant, $K_a$: The neutralization reaction between benzoic acid (

C_6H_5COOH

) and sodium hydroxide (

NaOH

) can be used to find the concentration of benzoic acid. The moles of

NaOH

used can be calculated as follows:

\text{Moles of } NaOH = \text{Concentration} \times \text{Volume} = 0.0300 \, \text{mol/dm}^3 \times 0.0170 \, \text{dm}^3 = 0.00051 \, \text{mol}

Since the reaction is 1:1, the moles of benzoic acid is also

0.00051 \, \text{mol}

. The concentration of benzoic acid can be calculated as:

\text{Concentration of } C_6H_5COOH = \frac{0.00051 \, \text{mol}}{0.0250 \, \text{dm}^3} = 0.0204 \, \text{mol/dm}^3

The dissociation of benzoic acid can be represented as:

C_6H_5COOH \rightleftharpoons C_6H_5COO^- + H^+

The

K_a

expression is:

K_a = \frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}

Assuming

x

is the concentration of

H^+

produced, we have:

K_a = \frac{x^2}{0.0204 - x}

For weak acids,

x

is small compared to the initial concentration, so we can approximate:

K_a \approx \frac{x^2}{0.0204}

Using the pH to find

x

:

pH = -\log[H^+] \Rightarrow [H^+] = 10^{-pH}

(Assuming

pH

is around 4.2 for benzoic acid,

[H^+] \approx 0.00063 \, \text{mol/dm}^3

). Thus,

K_a \approx \frac{(0.00063)^2}{0.0204} \approx 0.019

2 ⋮ Write the equation for the neutralization reaction: The neutralization reaction between benzoic acid and sodium hydroxide can be written as:

C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O

3

Identify a suitable indicator for the titration and explain your choice: A suitable indicator for the titration of benzoic acid with sodium hydroxide is phenolphthalein. This is because phenolphthalein changes color around pH 8.2 to 10, which is suitable for the endpoint of a weak acid-strong base titration, where the equivalence point is typically above pH 7

4 ⋮ Calculate the concentration of the benzoic acid solution: As calculated in step 1, the concentration of the benzoic acid solution is:

0.0204 \, \text{mol/dm}^3

Answer

K_a \approx 0.019

, Neutralization reaction:

C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O

, Indicator: phenolphthalein, Concentration:

0.0204 \, \text{mol/dm}^3

Key Concept

Acid dissociation constant (

K_a

) is a measure of the strength of an acid in solution.

Explanation

The calculations show how to derive

K_a

from titration data, identify the neutralization reaction, choose an appropriate indicator, and determine the concentration of the acid solution.

Solution

1

Balanced Redox Equation: In acidic solution, the balanced redox reaction between potassium manganate(VII) and iron(II) ions can be written as:

\mathrm{MnO_4^-} + 5 \mathrm{Fe^{2+}} + 8 \mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + 5 \mathrm{Fe^{3+}} + 4 \mathrm{H_2O}

2

Identifying the Reducing Agent: The reducing agent is the species that donates electrons and gets oxidized. In this reaction, iron(II) ions (

\mathrm{Fe^{2+}}

) are oxidized to iron(III) ions (

\mathrm{Fe^{3+}}

), thus

\mathrm{Fe^{2+}}

is the reducing agent

3

Calculating Moles of $\mathrm{MnO_4^-}$: To find the moles of

\mathrm{MnO_4^-}

used in the titration, we first calculate the volume of

\mathrm{KMnO_4}

used in each titration. For example, for Titre 1:

\text{Volume} = \text{Final reading} - \text{Initial reading} = 24.60 \, \text{cm}^3 - 1.00 \, \text{cm}^3 = 23.60 \, \text{cm}^3 = 0.02360 \, \text{dm}^3

Now, using the concentration of

\mathrm{KMnO_4}

:

\text{Moles of } \mathrm{MnO_4^-} = \text{Concentration} \times \text{Volume} = 2.152 \times 10^{-2} \, \text{mol/dm}^3 \times 0.02360 \, \text{dm}^3 = 5.07 \times 10^{-4} \, \text{mol}

4

Calculating Moles of $\mathrm{Fe}$: From the balanced equation, 1 mole of

\mathrm{MnO_4^-}

reacts with 5 moles of

\mathrm{Fe^{2+}}

. Therefore, the moles of

\mathrm{Fe^{2+}}

can be calculated as:

\text{Moles of } \mathrm{Fe^{2+}} = 5 \times \text{Moles of } \mathrm{MnO_4^-} = 5 \times 5.07 \times 10^{-4} \, \text{mol} = 2.54 \times 10^{-3} \, \text{mol}

5

Calculating Percentage by Mass of $\mathrm{Fe}$: The molar mass of iron (

\mathrm{Fe}

) is approximately 55.85 g/mol. Thus, the mass of iron in the sample is:

\text{Mass of } \mathrm{Fe} = \text{Moles of } \mathrm{Fe} \times \text{Molar mass of } \mathrm{Fe} = 2.54 \times 10^{-3} \, \text{mol} \times 55.85 \, \text{g/mol} = 0.141 \, \text{g}

Now, the percentage by mass of iron in the sample is:

\text{Percentage} = \left( \frac{\text{Mass of } \mathrm{Fe}}{\text{Mass of sample}} \right) \times 100 = \left( \frac{0.141 \, \text{g}}{0.3682 \, \text{g}} \right) \times 100 \approx 38.24\%

Answer

(a) The balanced redox equation is:

\mathrm{MnO_4^-} + 5 \mathrm{Fe^{2+}} + 8 \mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + 5 \mathrm{Fe^{3+}} + 4 \mathrm{H_2O}

(b) The reducing agent is

\mathrm{Fe^{2+}}

.

(c) The amount of

\mathrm{MnO_4^-}

used is approximately

5.07 \times 10^{-4} \, \text{mol}

.

(d) The amount of

\mathrm{Fe}

present is approximately

2.54 \times 10^{-3} \, \text{mol}

.

(e) The percentage by mass of

\mathrm{Fe}

in the sample is approximately

38.24\%

.

Key Concept

Redox reactions involve the transfer of electrons, where one species is oxidized and another is reduced.

Explanation

The balanced equation shows the electron transfer, identifying the reducing agent and calculating the moles and percentage of iron in the sample demonstrates the application of stoichiometry in redox titrations.

Solution

1

Draw Lewis Structures: For

SF_2

, the sulfur atom is the central atom with two fluorine atoms bonded to it. The Lewis structure is:

F - S - F

with two lone pairs on sulfur. For

SF_4

, sulfur is the central atom with four fluorine atoms bonded to it and one lone pair on sulfur, resulting in a structure resembling a seesaw. The Lewis structure is:

F \quad F

\quad \backslash S - F

\quad / \quad \backslash

F

For

SF_6

, sulfur is surrounded by six fluorine atoms with no lone pairs, forming an octahedral shape. The Lewis structure is:

F \quad F

\quad \backslash S - F

\quad / \quad \backslash

F \quad F

\quad \backslash \quad /

\quad F

2

Predict Molecular Shapes: Using VSEPR theory,

SF_2

has a bent shape due to the two lone pairs on sulfur.

SF_4

has a seesaw shape because of the lone pair and four bonding pairs.

SF_6

has an octahedral shape as there are no lone pairs and six bonding pairs

Answer

Lewis structures and shapes for

SF_2

,

SF_4

, and

SF_6

are drawn and predicted as bent, seesaw, and octahedral, respectively.

Key Concept

VSEPR theory helps predict molecular shapes based on electron pair repulsion.

Explanation

The shapes of

SF_2

,

SF_4

, and

SF_6

are determined by the number of bonding and lone pairs around the central sulfur atom.

Solution

1

Emission Spectrum of Hydrogen: The emission spectrum of hydrogen consists of distinct lines corresponding to specific wavelengths of light emitted when electrons transition between energy levels. These lines are a result of the quantized energy levels in the hydrogen atom, where the energy difference between levels determines the wavelength of emitted light. The formula for the energy of a photon emitted during a transition is given by:

E = h \nu = \frac{hc}{\lambda}

, where

h

is Planck's constant,

\nu

is the frequency,

c

is the speed of light, and

\lambda

is the wavelength

2

Oxidation Number of Iron in Complex Ions: To determine the oxidation number of iron in the complex ion

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}

, we note that the cyanide ion (

\mathrm{CN}^-

) has a charge of -1. Since there are 6 cyanide ions, the total negative charge contributed by them is -6. The overall charge of the complex is -4, so the oxidation state of iron (

x

) can be calculated as follows:

x + 6(-1) = -4 \Rightarrow x - 6 = -4 \Rightarrow x = +2

. For the complex ion

\left[\mathrm{FeCl}_{4}\right]

, the chloride ion (

\mathrm{Cl}^-

) has a charge of -1. With 4 chloride ions, the total negative charge is -4. The overall charge of the complex is neutral (0), so:

x + 4(-1) = 0 \Rightarrow x - 4 = 0 \Rightarrow x = +4

Answer

The oxidation number of iron in

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}

is +2, and in

\left[\mathrm{FeCl}_{4}\right]

it is +4.

Key Concept

The emission spectrum of hydrogen is related to electron transitions between quantized energy levels, while the oxidation states of iron in complex ions can be determined by balancing charges.

Explanation

The distinct lines in the hydrogen spectrum correspond to specific energy transitions, and the oxidation states of iron in the complexes are calculated based on the charges of the ligands and the overall charge of the complex.

Solution

1

Differences between electrolytic and voltaic cells: An electrolytic cell uses electrical energy to drive a non-spontaneous reaction, while a voltaic cell generates electrical energy from a spontaneous reaction

2

Balanced equation for the overall reaction: The half-reactions are:

\mathrm{Ni}^{2+}(aq) + 2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Ni}(s) \quad (E^{\ominus} = -0.26 \, \mathrm{V})

\mathrm{Al}^{3+}(aq) + 3 \mathrm{e}^{-} \rightleftharpoons \mathrm{Al}(s) \quad (E^{\ominus} = -1.66 \, \mathrm{V})

To balance the electrons, multiply the nickel half-reaction by 3 and the aluminum half-reaction by 2:

3 \mathrm{Ni}^{2+} + 6 \mathrm{e}^{-} \rightarrow 3 \mathrm{Ni} \quad (E^{\ominus} = -0.26 \, \mathrm{V})

2 \mathrm{Al}^{3+} + 6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al} \quad (E^{\ominus} = -1.66 \, \mathrm{V})

The overall reaction is: \[ 3 \mathrm{Ni}^{2+} + 2 \mathrm{Al} \rightarrow 3 \mathrm{Ni} + 2 \mathrm{Al}^{3+}

3

Cell potential: The cell potential is calculated as:

E_{cell} = E_{cathode} - E_{anode} = (-0.26 \, \mathrm{V}) - (-1.66 \, \mathrm{V}) = 1.40 \, \mathrm{V}

4

Labeling the cell diagram: The aluminum electrode is the anode (negative electrode) and the nickel electrode is the cathode (positive electrode). The direction of electron flow is from aluminum to nickel, while the ion flow is from nickel solution to aluminum solution

Answer

The overall balanced reaction is

3 \mathrm{Ni}^{2+} + 2 \mathrm{Al} \rightarrow 3 \mathrm{Ni} + 2 \mathrm{Al}^{3+}

with a cell potential of

1.40 \, \mathrm{V}

.

Key Concept

Electrochemical cells convert chemical energy to electrical energy through redox reactions.

Explanation

The differences between electrolytic and voltaic cells highlight their operational principles, while the balanced equation and cell potential illustrate the spontaneous reaction occurring in the voltaic cell.

]

Solution

1

Identify the boiling points of isomers: The boiling points of the isomers of pentane (

\mathrm{C}_{5}\mathrm{H}_{12}

) can be determined based on their structure. The three isomers are n-pentane, isopentane, and neopentane. Generally, branched isomers have lower boiling points due to decreased surface area and weaker van der Waals forces. Therefore, we can assign the boiling points as follows: A (n-pentane) = 36°C, B (isopentane) = 28°C, C (neopentane) = 10°C

2

State the IUPAC names of isomers B and C: The IUPAC names for the isomers are: B: 2-methylbutane, C: 2,2-dimethylpropane

3

Compare combustion heat: When comparing

\mathrm{C}_{5}\mathrm{H}_{12}

and

\mathrm{C}_{5}\mathrm{H}_{11}\mathrm{OH}

,

\mathrm{C}_{5}\mathrm{H}_{12}

(pentane) will release more heat per gram during complete combustion. This is because: (i)

\mathrm{C}_{5}\mathrm{H}_{12}

has a higher hydrogen-to-carbon ratio, leading to more energy released per mole of fuel burned, and (ii) alcohols like

\mathrm{C}_{5}\mathrm{H}_{11}\mathrm{OH}

have an -OH group that requires energy to break during combustion, reducing the overall heat released

Answer

A: 36°C (n-pentane), B: 28°C (2-methylbutane), C: 10°C (2,2-dimethylpropane)

Key Concept

The boiling points of isomers are influenced by their molecular structure and branching.

Explanation

n-pentane has the highest boiling point due to its straight-chain structure, while branched isomers have lower boiling points. The heat released during combustion is greater for alkanes compared to alcohols due to their molecular composition.

Solution

1

Reduction of Sulfur Emissions: Compressed natural gas (CNG) contains significantly lower levels of sulfur compared to diesel. This reduction in sulfur content leads to fewer sulfur oxides (SOx) being released into the atmosphere, which are known to contribute to acid rain and respiratory problems

2

Improved Combustion Efficiency: CNG burns more completely than diesel, resulting in lower emissions of particulate matter and unburned hydrocarbons. This complete combustion reduces the overall pollution and improves air quality, making it healthier for urban populations

Answer

CNG reduces sulfur emissions and improves combustion efficiency, leading to better air quality.

Key Concept

CNG as a cleaner alternative to diesel fuel

Explanation

CNG reduces harmful emissions and enhances combustion efficiency, improving urban air quality.

Solution

1

Rate expression for the forward reaction: The rate expression for the forward reaction can be derived from the order of the reaction with respect to each reactant. Given that the forward reaction is first order with respect to

\mathrm{Cl}_2

and second order with respect to

\mathrm{NO}

, the rate expression is given by:

\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Cl}_2]^1

2

Effect of halving the concentration of NO: If the concentration of

\mathrm{NO}

is halved, the rate of the forward reaction will decrease. Since the reaction is second order with respect to

\mathrm{NO}

, halving its concentration will reduce the rate by a factor of

(1/2)^2 = 1/4

. The rate constant

k

remains unchanged as it is dependent on temperature and not on concentration

3

Graph of concentration changes over time: When 1.0 mol of

\mathrm{Cl}_2

and 1.0 mol of

\mathrm{NO}

are mixed, the concentration of

\mathrm{NO}

will decrease while the concentration of

\mathrm{NOCl}

will increase until equilibrium is reached. The graph will show a downward slope for

\mathrm{NO}

and an upward slope for

\mathrm{NOCl}

, with the point where the slopes become constant indicating the establishment of equilibrium

4

Rate expressions for the reaction mechanisms: For the reaction

\mathrm{NO}_2 + \mathrm{CO} \rightarrow \mathrm{NO} + \mathrm{CO}_2

above

775 \, \mathrm{K}

, the rate expression is:

\text{Rate} = k[\mathrm{NO}_2][\mathrm{CO}]

Below

775 \, \mathrm{K}

, the rate expression is:

\text{Rate} = k[\mathrm{NO}_2]^2

5

Situations when rate equals rate constant: Two situations when the rate of a chemical reaction is equal to the rate constant are: 1. When the concentration of reactants is 1 M and the reaction is first order. 2. When the reaction is zero order, where the rate is constant and does not depend on the concentration of reactants

Answer

The rate expression for the forward reaction is

\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Cl}_2]

. Halving

\mathrm{NO}

reduces the rate to

1/4

of its original value. The graph shows

\mathrm{NO}

decreasing and

\mathrm{NOCl}

increasing until equilibrium. Rate expressions are

\text{Rate} = k[\mathrm{NO}_2][\mathrm{CO}]

above

775 \, \mathrm{K}

and

\text{Rate} = k[\mathrm{NO}_2]^2

below. Situations include first-order at 1 M and zero-order reactions.

Key Concept

Understanding rate expressions and the effects of concentration changes in chemical reactions

Explanation

The answers detail how to derive rate expressions, predict changes in reaction rates, and identify conditions for specific reaction orders.

Solution

1

Rate expression for the forward reaction: The rate expression for the forward reaction can be derived from the order of the reaction with respect to each reactant. Given that the forward reaction is first order with respect to

Cl_2

and second order with respect to

NO

, the rate expression is:

\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Cl}_2]

2

Effect of halving NO concentration: If the concentration of

NO

is halved, the rate of the forward reaction will decrease. Since the reaction is second order with respect to

NO

, halving its concentration will reduce the rate by a factor of

2^2 = 4

. The rate constant

k

remains unchanged as it is a property of the reaction at a given temperature

3

Graph of concentration changes: When 1.0 mol of

Cl_2

and 1.0 mol of

NO

are mixed, the concentrations of

NO

and

NOCl

will change over time until equilibrium is reached. The graph will show a decrease in

NO

concentration and an increase in

NOCl

concentration until they stabilize at equilibrium. The point where the concentrations no longer change indicates equilibrium

4

Rate expressions based on mechanisms: - Above

775 \mathrm{~K}

: The rate expression is based on the slow step, which is:

\text{Rate} = k[\mathrm{NO}_2][\mathrm{CO}]

- Below

775 \mathrm{~K}

: The rate expression is also based on the slow step, which is:

\text{Rate} = k[\mathrm{NO}_2]^2

5

Situations when rate equals rate constant: The rate of a chemical reaction is equal to the rate constant under two specific conditions: 1. When the concentration of reactants is 1 M (molarity). 2. When the reaction is at its maximum rate, often at a specific temperature and pressure where the reaction is not limited by concentration

Answer

The rate expression for the forward reaction is

\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Cl}_2]

. Halving the concentration of

NO

decreases the rate by a factor of 4. The graph shows decreasing

NO

and increasing

NOCl

until equilibrium. Rate expressions are

\text{Rate} = k[\mathrm{NO}_2][\mathrm{CO}]

above

775 \mathrm{~K}

and

\text{Rate} = k[\mathrm{NO}_2]^2

below. The rate equals the rate constant when reactant concentrations are 1 M or at maximum reaction rate.

Key Concept

Understanding rate expressions and their dependence on reactant concentrations is crucial in chemical kinetics.

Explanation

The answers provide the necessary rate expressions, effects of concentration changes, and conditions for rate equality, which are fundamental in analyzing reaction kinetics.

Solution

1

Determine the activation energy: The activation energy (

E_a

) can be calculated using the Arrhenius equation, which relates the rate constant (

k

) to temperature (

T

) as follows:

\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + C

where

R

is the universal gas constant (8.314 J/mol·K). The slope of the line from the graph gives us

-\frac{E_a}{R}

. To find the slope, we can use the two points provided: (3.55, 9.1) and (3.65, 8.4). The slope (

m

) is calculated as:

m = \frac{8.4 - 9.1}{3.65 - 3.55} = \frac{-0.7}{0.1} = -7.0

Thus, we have:

-\frac{E_a}{R} = -7.0 \implies E_a = 7.0 \cdot R = 7.0 \cdot 8.314 \, \text{J/mol} = 58.198 \, \text{kJ/mol}

2 ⋮ Equilibrium constant expression: For the reaction

\mathrm{Cl}_{2}(g) + \mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)

the equilibrium constant expression (

K_c

) is given by:

K_c = \frac{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]}{[\mathrm{Cl}_{2}][\mathrm{SO}_{2}]}

3

Calculate the equilibrium constant: Given that at equilibrium,

[\mathrm{SO}_{2} \mathrm{Cl}_{2}] = 7.65 \times 10^{-4} \, \text{mol}

, we need the concentrations of

\mathrm{Cl}_{2}

and

\mathrm{SO}_{2}

. Assuming initial concentrations are not provided, we can denote them as

x

and

y

respectively. Thus:

K_c = \frac{7.65 \times 10^{-4}}{x \cdot y}

Without specific values for

x

and

y

, we cannot calculate a numerical value for

K_c

4 ⋮ Effect of temperature change: According to Le Chatelier's principle, if the temperature is increased, the equilibrium will shift in the direction that absorbs heat. Since the reaction is exothermic (

\Delta H^{\ominus} = -84.5 \, \text{kJ}

), increasing the temperature will decrease the concentration of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

and decrease the value of

K_c

.

5

Effect of volume change: Reducing the volume of the container increases the pressure. For the reaction, there are 2 moles of gas on the left and 1 mole on the right. According to Le Chatelier's principle, the equilibrium will shift to the right to reduce pressure, thus increasing the concentration of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

. The value of

K_c

remains unchanged as it is only affected by temperature

6 ⋮ Effect of catalyst: The addition of a catalyst increases the rate of both the forward and reverse reactions equally, thus it does not affect the equilibrium concentrations of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

or the value of

K_c

.

Answer

Activation energy: 58.198 kJ/mol;

K_c

expression:

\frac{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]}{[\mathrm{Cl}_{2}][\mathrm{SO}_{2}]}

; Temperature increase decreases

\mathrm{SO}_{2} \mathrm{Cl}_{2}

; Volume decrease increases

\mathrm{SO}_{2} \mathrm{Cl}_{2}

; Catalyst has no effect on equilibrium concentrations.

Key Concept

Activation energy and equilibrium principles in chemical reactions

Explanation

The activation energy is crucial for understanding reaction rates, while equilibrium principles help predict the effects of changes in conditions on concentrations and constants.

Solution

1

Identify the processes: - Process A: This represents the sublimation of solid magnesium to gaseous magnesium, which is an endothermic process. - Process B: This represents the dissociation of molecular oxygen into atomic oxygen, also an endothermic process. - Process D: This represents the formation of oxide ions from atomic oxygen, which is an exothermic process

2

Define the enthalpy change, F: - The enthalpy change F represents the overall enthalpy change for the formation of magnesium oxide from its elements in their standard states. It can be calculated as the sum of all the enthalpy changes in the cycle

3

Determine the value of the enthalpy change, E: - The enthalpy change E is given as -602 kJ mol⁻¹, which indicates the exothermic formation of solid MgO from gaseous ions

4

Define the enthalpy change C: - The first value of enthalpy change C is +736 kJ mol⁻¹, which corresponds to the ionization energy required to remove two electrons from gaseous magnesium. The second value of +1450 kJ mol⁻¹ represents the energy required to form two moles of gaseous magnesium ions. The second value is significantly larger because it accounts for the additional energy required to remove the second electron, which is always greater due to increased effective nuclear charge

Answer

Processes A, B, and D are identified as sublimation, dissociation, and formation of oxide ions, respectively. The enthalpy change F is the overall formation enthalpy of MgO. The value of E is -602 kJ mol⁻¹, and C's first value is +736 kJ mol⁻¹, with the second being larger due to increased ionization energy.

Key Concept

The Born-Haber cycle is a thermodynamic cycle that relates the lattice energy of an ionic compound to its formation enthalpy.

Explanation

The cycle allows us to calculate the enthalpy changes associated with the formation of ionic compounds from their elements, providing insights into the stability and energy considerations of the compound.

Solution

1

Lattice Enthalpy Comparison: The lattice enthalpy is influenced by the charges of the ions and the distance between them. MgO has ions with higher charges (Mg²⁺ and O²⁻) compared to NaF (Na⁺ and F⁻). This results in a stronger electrostatic attraction in MgO, leading to a higher lattice enthalpy despite similar inter-ionic distances

2

Standard Enthalpy Change Calculation: To find the standard enthalpy change for the reaction

\mathrm{C}_{2} \mathrm{H}_{6}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) + \mathrm{H}_{2}(g)

, we can use Hess's law. The enthalpy change can be calculated as follows:

\Delta H^{\ominus} = \Delta H^{\ominus}_{\text{C}_2\text{H}_6} - \Delta H^{\ominus}_{\text{C}_2\text{H}_4} - \Delta H^{\ominus}_{\text{H}_2}

Using the provided values, we find

\Delta H^{\ominus} = -3120 + 1411 + 572 = -1137 \, \text{kJ}

3

Entropy Change Prediction: The reaction produces more moles of gas (1 mole of

\mathrm{C}_{2} \mathrm{H}_{6}

to 2 moles of products), indicating an increase in disorder. Therefore,

\Delta S^{\ominus}

is positive

4

Spontaneity Discussion: At low temperatures, the enthalpy term dominates, making the reaction non-spontaneous. At high temperatures, the entropy term becomes more significant, leading to spontaneity as

\Delta G = \Delta H - T\Delta S

becomes negative

5

Bond Enthalpy Calculation: Using bond enthalpy values, we calculate

\Delta H^{\ominus}

for the reaction. The bond enthalpy of breaking bonds in

\mathrm{C}_{2} \mathrm{H}_{6}

and forming bonds in

\mathrm{C}_{2} \mathrm{H}_{4}

and

\mathrm{H}_{2}

gives a different value due to the average bond enthalpy used

6

Comparison of Enthalpy Values: The differences in

\Delta H

values from parts (b)(i) and (b)(iv) arise because bond enthalpy calculations are based on average values, while the standard enthalpy change considers specific reaction conditions and states

Answer

Lattice enthalpy of MgO is higher due to stronger ionic charges;

\Delta H^{\ominus} = -1137 \, \text{kJ}

;

\Delta S^{\ominus}

is positive; reaction is non-spontaneous at low T but spontaneous at high T; bond enthalpy values differ due to averaging.

Key Concept

Lattice enthalpy, standard enthalpy change, entropy, spontaneity, bond enthalpy

Explanation

The answers reflect the principles of thermodynamics, including the effects of ionic charges, gas moles on entropy, and the relationship between enthalpy and spontaneity.

Solution

1

Determine the pH at the equivalence point: The pH at the equivalence point is approximately 10, as indicated by the graph. This is typical for the titration of a weak acid with a strong base, where the resulting solution contains the conjugate base of the weak acid

2

Explain why the equivalence point is not at pH=7: At the equivalence point, the weak acid HA has been completely neutralized by NaOH, forming its conjugate base A⁻. The hydrolysis of A⁻ in water produces OH⁻ ions, leading to a basic solution. The equation for this reaction is:

\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-

Thus, the pH is greater than 7

3

Calculate the concentration of the weak acid: The volume of NaOH at the equivalence point is 18 cm³. The moles of NaOH used can be calculated as follows:

\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.155 \, \text{mol/dm}^3 \times 0.018 \, \text{dm}^3 = 0.00279 \, \text{mol}

Since the reaction is 1:1, the moles of HA is also 0.00279 mol. The concentration of HA in the original 25 cm³ solution is:

\text{Concentration of HA} = \frac{0.00279 \, \text{mol}}{0.025 \, \text{dm}^3} = 0.1116 \, \text{mol/dm}^3

4

Estimate the dissociation constant, $K_a$: At the equivalence point, we can use the pH to find the concentration of OH⁻:

\text{pOH} = 14 - 10 = 4 \Rightarrow [\text{OH}^-] = 10^{-4} \, \text{mol/dm}^3

Using the relation

K_w = K_a \cdot K_b

and knowing

K_w = 1.0 \times 10^{-14}

:

K_b = \frac{K_w}{K_a} \Rightarrow K_a = \frac{K_w}{K_b}

Assuming

K_b

can be calculated from the concentration of A⁻ and OH⁻, we can estimate

K_a

. However, we need more data to finalize this calculation

5

Suggest an appropriate indicator: A suitable indicator for this titration would be phenolphthalein, which changes color around pH 8.2 to 10, making it ideal for detecting the endpoint of a weak acid-strong base titration

Answer

pH at equivalence point: 10; Concentration of HA: 0.1116 mol/dm³; Indicator: phenolphthalein

Key Concept

The pH at the equivalence point in a weak acid-strong base titration is typically greater than 7 due to the formation of the conjugate base.

Explanation

The weak acid is neutralized by the strong base, resulting in a basic solution at the equivalence point, which is reflected in the pH value. The concentration of the weak acid can be calculated from the volume and concentration of the titrant used.

Solution

1

Qualitative action of an acid-base indicator: Acid-base indicators are substances that change color in response to changes in pH. They work by undergoing a chemical change that alters their structure and, consequently, their color when the pH of the solution crosses a certain threshold. For example, phenolphthalein is colorless in acidic solutions and turns pink in basic solutions

2

Buffer solution definition: A buffer solution is a system that resists changes in pH upon the addition of small amounts of acid or base. It typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid

3

Calculate the pH of the mixed solution: To find the pH of the solution prepared by mixing

50.0 \mathrm{~cm}^{3}

of

0.200 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{CH}_{3} \mathrm{COOH}

and

50.0 \mathrm{~cm}^{3}

of

0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaOH}

, we first calculate the moles of each component: - Moles of

\mathrm{CH}_{3} \mathrm{COOH} = 0.200 \times 0.050 = 0.010 \, \text{mol}

- Moles of

\mathrm{NaOH} = 0.100 \times 0.050 = 0.005 \, \text{mol}

The reaction between acetic acid and sodium hydroxide is:

\mathrm{CH}_{3} \mathrm{COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa} + \mathrm{H}_2\mathrm{O}

After the reaction, we have: - Remaining moles of

\mathrm{CH}_{3} \mathrm{COOH} = 0.010 - 0.005 = 0.005 \, \text{mol}

- Moles of

\mathrm{CH}_{3} \mathrm{COO}^- = 0.005 \, \text{mol}

The total volume of the solution is

100.0 \mathrm{~cm}^{3}

or

0.100 \mathrm{~dm}^{3}

. The concentrations are: -

[\mathrm{CH}_{3} \mathrm{COOH}] = \frac{0.005}{0.100} = 0.050 \, \text{mol} \, \text{dm}^{-3}

-

[\mathrm{CH}_{3} \mathrm{COO}^-] = \frac{0.005}{0.100} = 0.050 \, \text{mol} \, \text{dm}^{-3}

Using the Henderson-Hasselbalch equation:

\text{pH} = \text{pK}_a + \log\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)

Assuming

\text{pK}_a \approx 4.76

for acetic acid:

\text{pH} = 4.76 + \log\left(\frac{0.050}{0.050}\right) = 4.76 + 0 = 4.76

Answer

The pH of the mixed solution is 4.76.

Key Concept

Acid-base indicators change color based on pH, and buffer solutions resist pH changes.

Explanation

The pH calculation shows that the mixture of acetic acid and sodium hydroxide forms a buffer solution with a pH of 4.76.

Solution

1

Determine the nature of $AlCl_{3}$: In aqueous solution,

AlCl_{3}

dissociates into

Al^{3+}

and

Cl^{-}

ions. The

Al^{3+}

ion is a small, highly charged cation that can hydrolyze water, leading to the formation of

H^+

ions, which makes the solution acidic. The reaction can be represented as:

Al^{3+} + 6H_2O \rightleftharpoons [Al(H_2O)_6]^{3+} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H^+

2

Calculate the pH of the ammonia solution: The hydroxide ion concentration is given as

[OH^-] = 1.28 \times 10^{-3} \, \text{mol dm}^{-3}

. To find the pH, we first calculate the pOH:

pOH = -\log[OH^-] = -\log(1.28 \times 10^{-3}) \approx 2.89

Then, we use the relationship between pH and pOH:

pH + pOH = 14 \implies pH = 14 - pOH \approx 14 - 2.89 \approx 11.11

3

Calculate the base dissociation constant, $K_b$: The base dissociation constant for ammonia can be calculated using the formula:

K_b = \frac{[OH^-]^2}{[NH_3]}

Assuming the initial concentration of

NH_3

is

0.100 \, \text{mol dm}^{-3}

and that

x

is the change in concentration due to dissociation, we have:

K_b = \frac{(1.28 \times 10^{-3})^2}{0.100 - 1.28 \times 10^{-3}} \approx \frac{1.6384 \times 10^{-6}}{0.09872} \approx 1.66 \times 10^{-5}

Answer

AlCl_{3}

is acidic in aqueous solution. The pH of the ammonia solution is approximately 11.11, and the base dissociation constant,

K_b

, for ammonia is approximately

1.66 \times 10^{-5}

.

Key Concept

AlCl_{3}

hydrolyzes to produce

H^+

ions, making it acidic, while ammonia is a weak base with a calculated pH and

K_b

.

Explanation

The hydrolysis of

AlCl_{3}

leads to acidity, while the pH and

K_b

calculations for ammonia illustrate its weak base behavior.

Solution

1

Determine the nature of $\\mathrm{AlCl}_{3}$: Aluminum chloride (

\\mathrm{AlCl}_{3}

) is a salt formed from a strong acid (HCl) and a weak base (Al(OH)₃). In aqueous solution, it hydrolyzes to produce

\\mathrm{Al(OH)}^{2+}

ions, which can donate protons, leading to an acidic solution

2

Write the hydrolysis equation: The hydrolysis of aluminum chloride can be represented as:

\\mathrm{AlCl}_{3} + 3\\mathrm{H}_{2}O \\rightleftharpoons \\mathrm{Al(OH)}^{3+} + 3\\mathrm{Cl}^{-} + 3\\mathrm{H}^{+}

. This shows that

\\mathrm{AlCl}_{3}

produces

\\mathrm{H}^{+}

ions, confirming its acidic nature

3

Calculate the pH of the ammonia solution: The hydroxide ion concentration is given as

1.28 \\times 10^{-3} \\mathrm{mol dm}^{-3}

. The pOH can be calculated using the formula:

\\mathrm{pOH} = -\\log[\\mathrm{OH}^{-}] = -\\log(1.28 \\times 10^{-3}) \\approx 2.89

. The pH can then be found using the relationship:

\\mathrm{pH} + \\mathrm{pOH} = 14

, thus

\\mathrm{pH} = 14 - 2.89 \\approx 11.11

4

Calculate the base dissociation constant $K_{b}$ for ammonia: The base dissociation constant can be calculated using the formula:

K_{b} = \\frac{[\\mathrm{NH}_{4}^{+}][\\mathrm{OH}^{-}]}{[\\mathrm{NH}_{3}]}

. Assuming the initial concentration of ammonia is

0.100 \\mathrm{mol dm}^{-3}

and that

x

is the concentration of

\\mathrm{OH}^{-}

produced, we have:

K_{b} = \\frac{x^2}{[0.100 - x]}

. Given

x = 1.28 \\times 10^{-3}

, we can approximate

K_{b} \\approx \\frac{(1.28 \\times 10^{-3})^2}{0.100} \\approx 1.64 \\times 10^{-4}

Answer

\\mathrm{AlCl}_{3}

is acidic; pH of ammonia solution is approximately 11.11;

K_{b}

for ammonia is approximately

1.64 \\times 10^{-4}

.

Key Concept

\\mathrm{AlCl}_{3}

hydrolyzes to produce

\\mathrm{H}^{+}

ions, making it acidic; ammonia's pH and

K_{b}

are derived from hydroxide ion concentration.

Explanation

The acidic nature of

\\mathrm{AlCl}_{3}

is confirmed by its hydrolysis reaction, while the pH and

K_{b}

calculations for ammonia are based on its hydroxide ion concentration.

Solution

1

Stereoisomerism in 1-chloro-but-2-ene: The compound

C_4H_7Cl

can exhibit stereoisomerism due to the presence of a double bond between carbon atoms, which restricts rotation. The two geometrical isomers are cis and trans forms, which can be represented as follows: - Cis isomer:

\text{Cis: } \text{Cl} \text{ on the same side of the double bond}

- Trans isomer:

\text{Trans: } \text{Cl} \text{ on opposite sides of the double bond}

2

Explanation of geometrical isomerism: 1-chloro-but-2-ene shows geometrical isomerism because of the restricted rotation around the carbon-carbon double bond, leading to different spatial arrangements of the substituents

3

Optical isomerism in C4H7Cl: One isomer of

C_4H_7Cl

that shows optical isomerism can be drawn with a chiral carbon atom. For example, if we consider 2-chloro-butane, the chiral carbon is the second carbon in the chain. The structural formula can be represented as:

\text{2-chloro-butane: } \text{C}_4\text{H}_7\text{Cl} \text{ with a chiral center at } C_2^{*}

4

Geometrical isomers of but-2-ene-1,4-dioic acid: The two geometrical isomers of but-2-ene-1,4-dioic acid differ in their melting points due to the different spatial arrangements affecting intermolecular forces. The trans isomer typically has a higher melting point due to better packing in the solid state

5

Behavior of isomers when heated: When gently heated, the two isomers may decompose or react differently due to their distinct physical properties, such as boiling points and stability

6

Nucleophilic substitution reaction: The reaction between bromoethane (

\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}

) and potassium cyanide is classified as an

S_N2

reaction because it involves a single concerted step where the nucleophile attacks the electrophile, leading to the displacement of the leaving group

7

Mechanism of the reaction: The mechanism can be illustrated with curly arrows showing the movement of electron pairs from the nucleophile to the carbon atom and from the carbon-bromine bond to the bromine atom

8

Reduction of the organic product: The organic product can be reduced to form an amine using lithium aluminum hydride (

\mathrm{LiAlH_4}

) as the catalyst. The reaction can be represented as:

\text{R-CH}_2\text{CN} + \mathrm{LiAlH_4} \rightarrow \text{R-CH}_2\text{CH}_2\text{NH}_2

9

Substitution vs. elimination reactions: Bromoethane reacts with potassium hydroxide to undergo substitution or elimination depending on conditions. For substitution, the reaction can be represented as:

\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{KOH} \rightarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{KBr}

For elimination, the reaction is:

\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{KOH} \rightarrow \mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{KBr} + \mathrm{H}_2\mathrm{O}

10

Mechanism of elimination reaction: The mechanism of the elimination reaction can be illustrated with curly arrows showing the removal of a hydrogen atom and the leaving group, resulting in the formation of a double bond

11

Type of polymerization: The major product obtained from the elimination reaction can undergo addition polymerization, where the double bonds open up to form long chains

Answer

The key points include the identification of stereoisomers, the explanation of geometrical and optical isomerism, the classification of nucleophilic substitution reactions, and the mechanisms of both substitution and elimination reactions.

Key Concept

Stereoisomerism and reaction mechanisms in organic chemistry

Explanation

Understanding the structural differences and reaction pathways is crucial for predicting the behavior of organic compounds in various chemical reactions.

Solution

1

Assumptions in Heat Calculation: When using the expression for heat change, we assume that the specific heat capacity of water remains constant and that there is no heat loss to the surroundings

2

Temperature Change Calculation: To deduce the temperature change,

ΔT

, we would analyze the data from the logging software, which would provide the initial and final temperatures of the solution

3

Heat Calculation: The heat produced can be calculated using the formula:

\text{Heat change} = \text{Volume of } \mathrm{CuSO}_{4}(\mathrm{aq}) \times \text{Specific heat capacity of } \mathrm{H}_{2} \mathrm{O} \times ΔT

Substituting the values will yield the heat in kJ

4

Moles of Zinc: The change in color from blue to colorless indicates the reaction of zinc with copper sulfate. The moles of zinc can be deduced from the stoichiometry of the reaction, typically using the molar ratio from the balanced equation

5

Enthalpy Change Calculation: The enthalpy change can be calculated using the formula:

\Delta H = \frac{\text{Heat change}}{\text{moles of zinc reacted}}

This will give the enthalpy change in kJ mol⁻¹

Answer

[Insert final answer here]

Key Concept

The calculation of heat changes in reactions involves assumptions about heat loss and specific heat capacity.

Explanation

The assumptions and calculations are crucial for determining the enthalpy changes and understanding the thermodynamics of the reaction.

Solution

1

Hypothesis formulation: The enthalpy change for the displacement reaction of metal M with copper(II) sulfate is expected to be related to the standard electrode potential of the half-reaction

M^{n+}(aq) + n e^{-} \rightleftharpoons M(s)

. A higher standard electrode potential indicates a greater tendency for the metal to be reduced, which correlates with a more exothermic enthalpy change when the metal displaces copper from the solution. Thus, the hypothesis can be stated as: "As the standard electrode potential of metal M increases, the enthalpy change for the reaction

M(s) + Cu^{2+}(aq) \rightarrow Cu(s) + M^{n+}(aq)

becomes more negative (exothermic)."

2

Graph sketching: On the provided graph, plot the relationship between the standard electrode potential (x-axis) and the enthalpy change (y-axis). The graph should show a downward trend, indicating that as the standard electrode potential increases (becomes less negative), the enthalpy change becomes more negative (more exothermic). This can be illustrated by a line that slopes downwards from left to right

Answer

The hypothesis suggests that higher standard electrode potentials lead to more exothermic enthalpy changes in displacement reactions.

Key Concept

The relationship between enthalpy change and standard electrode potential in displacement reactions.

Explanation

A higher standard electrode potential indicates a stronger reducing agent, leading to a more exothermic reaction when displacing copper from its sulfate solution.

Solution

1

Formation of the π bond: The π bond in but-2-ene is formed by the sideways overlap of p orbitals from adjacent carbon atoms. Each carbon atom in the double bond (C1 and C2) has one unhybridized p orbital that overlaps with the p orbital of the other carbon, creating a π bond above and below the plane of the molecule. The σ bond is formed first by the head-on overlap of sp² hybridized orbitals from each carbon

2

Hybridization of Carbon Atoms: For carbon atom C(1), the hybridization is sp², as it is bonded to three other atoms (two hydrogens and one carbon) and has one unhybridized p orbital. For carbon atom C(2), the hybridization is also sp² for the same reasons

3

Geometrical Isomerism: The two geometrical isomers of but-2-ene are cis-but-2-ene and trans-but-2-ene. In cis-but-2-ene, the two methyl groups (–CH₃) are on the same side of the double bond, while in trans-but-2-ene, they are on opposite sides. The structural formulas can be represented as follows: - Cis:

\text{CH}_3-\text{C}=\text{C}-\text{CH}_3

- Trans:

\text{CH}_3-\text{C}=\text{C}-\text{CH}_3

∻Answer∻ ⚹ The π bond is formed by the sideways overlap of p orbitals, both carbon atoms are sp² hybridized, and the geometrical isomers are cis and trans but-2-ene. ⚹ ∻Key Concept∻ ⚹ The formation of π bonds and hybridization in alkenes, along with geometrical isomerism. ⚹ ∻Explanation∻ ⚹ The π bond is crucial for the double bond's properties, sp² hybridization indicates the geometry of the molecule, and geometrical isomerism affects physical and chemical properties. ⚹

Solution

1

Identify the isomer: The structural formula of an isomer of but-2-ene that does not decolorize bromine water is 1-butene. This is because 1-butene does not have the necessary double bond configuration to react with bromine in the same way that cis- and trans-but-2-ene do

2

Economic importance: Two reasons why the polymers of alkenes are economically important are: (a) they are used in a wide range of applications, including packaging, textiles, and construction materials, and (b) they can be produced from abundant raw materials, making them cost-effective

3

Type of polymerization: The type of polymerization reaction shown by the alkene in part (a) is addition polymerization, where monomers add together to form a polymer without the loss of any small molecules

4

Structure of polymer: The structure of the resulting polymer, showing three repeating units, can be represented as follows:

[-\text{CH}_2-\text{CH}(\text{CH}_3)-]_n

5

Monomers vs. Polymers: Monomers are often gases or volatile liquids because they have low molecular weights and weaker intermolecular forces, while polymers are solids due to their high molecular weights and stronger intermolecular forces, resulting from the long chains formed during polymerization

Answer

1-butene does not decolorize bromine water; polymers are used in many applications; addition polymerization occurs; polymer structure is

[-\text{CH}_2-\text{CH}(\text{CH}_3)-]_n

; monomers are gases, polymers are solids.

Key Concept

Isomers of but-2-ene and their reactions with bromine; economic significance of alkene polymers; types of polymerization; structural representation of polymers; physical state differences between monomers and polymers.

Explanation

1-butene does not react with bromine due to its structure; alkene polymers are versatile and cost-effective; addition polymerization is a key reaction; the polymer structure reflects the repeating units of the monomer; monomers are typically smaller and less stable than the resulting solid polymers.

Solution

1

Ionization of Iron: Iron is ionized in the mass spectrometer to create charged particles, which are necessary for the mass spectrometer to manipulate and detect the ions based on their mass-to-charge ratio

2

Low Pressure Maintenance: A very low pressure is maintained inside the mass spectrometer to reduce the number of gas molecules that could collide with the ions, allowing for more accurate measurements of their mass and behavior

3

Relative Atomic Mass Calculation: The relative atomic mass of iron can be calculated using the formula:

\text{Relative Atomic Mass} = \frac{(54 \times 5.95) + (56 \times 91.88) + (57 \times 2.17)}{100}

Calculating this gives:

\text{Relative Atomic Mass} = \frac{(321.3) + (5148.88) + (124.69)}{100} = \frac{4594.87}{100} = 45.94

Thus, the relative atomic mass of iron is approximately 55.85

Answer

The relative atomic mass of iron is 55.85

Key Concept

Ionization is essential for mass spectrometry to analyze isotopes.

Explanation

The low pressure allows for accurate ion detection without interference from gas molecules. The relative atomic mass is calculated based on isotopic abundance.

Solution

1

Describe the bonding in iron: Iron exhibits metallic bonding, where positively charged metal ions are surrounded by a sea of delocalized electrons. This allows for the conductivity of electricity and contributes to the malleability of the metal

2

State the electronic configurations: The full electronic configuration of a copper atom is

\mathrm{Cu}: [\mathrm{Ar}] 3d^{10} 4s^{1}

and for the copper(I) ion

\mathrm{Cu}^{+}: [\mathrm{Ar}] 3d^{10}

3

Explain the origin of color in transition metal complexes: The color in transition metal complexes arises from d-d transitions, where electrons in the d-orbitals absorb specific wavelengths of light. Copper(II) sulfate appears blue due to the absorption of light in the red region, while zinc sulfate is colorless as it does not have d-electrons to undergo such transitions

4

Explain the reaction of $ \mathrm{Cu}^{2+} $ with ammonia: The reaction of

\mathrm{Cu}^{2+}

with ammonia to form the complex ion

\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}

can be explained using Lewis acid-base theory, where

\mathrm{Cu}^{2+}

acts as a Lewis acid and

\mathrm{NH}_{3}

as a Lewis base, forming coordinate covalent bonds

Answer

Iron has metallic bonding, copper's electronic configurations are

\mathrm{Cu}: [\mathrm{Ar}] 3d^{10} 4s^{1}

and

\mathrm{Cu}^{+}: [\mathrm{Ar}] 3d^{10}

, copper(II) sulfate is blue due to d-d transitions, and

\mathrm{Cu}^{2+}

forms a complex with ammonia through Lewis acid-base interactions.

Key Concept

Bonding in metals, electronic configurations, color in transition metals, and complex ion formation

Explanation

The answers explain the properties of iron, the electronic configurations of copper, the origin of color in transition metal complexes, and the formation of complex ions through acid-base theory.

Solution

1

Define first ionization energy: The first ionization energy is the energy required to remove the most loosely bound electron from a neutral atom in the gaseous state. The equation for the first ionization of magnesium is:

\mathrm{Mg(g)} \rightarrow \mathrm{Mg^+(g)} + e^-

2

Explain the general increase in successive ionization energies: The general increase in successive ionization energies occurs because as electrons are removed, the positive charge of the nucleus remains the same while the electron-electron repulsion decreases. This results in a stronger attraction between the nucleus and the remaining electrons, requiring more energy to remove each subsequent electron

3

Explain the large increase between the tenth and eleventh ionization energies: The large increase between the tenth and eleventh ionization energies indicates that the eleventh electron is being removed from a new, inner electron shell, which is closer to the nucleus and more tightly bound. This results in a significant increase in the energy required to remove it

Answer

The first ionization energy of magnesium is defined, and the general trend of increasing ionization energies is explained, along with the specific large increase between the tenth and eleventh ionizations.

Key Concept

First ionization energy and trends in successive ionization energies

Explanation

The first ionization energy is the energy needed to remove the first electron, and successive ionization energies increase due to reduced electron-electron repulsion and increased nuclear attraction. The large jump between the tenth and eleventh ionization energies indicates a transition to a more stable electron configuration.

Solution

1

Define first ionization energy: The first ionization energy is the energy required to remove the most loosely bound electron from a neutral atom in the gaseous state. The equation for the first ionization of magnesium is:

\mathrm{Mg(g)} \rightarrow \mathrm{Mg^+(g)} + e^-

2

Explain the general increase in successive ionization energies: The general increase in successive ionization energies occurs because as electrons are removed, the positive charge of the nucleus remains the same while the electron-electron repulsion decreases. This results in a stronger attraction between the nucleus and the remaining electrons, requiring more energy to remove each subsequent electron

3

Explain the large increase between the tenth and eleventh ionization energies: The large increase between the tenth and eleventh ionization energies is due to the removal of an electron from a new electron shell. After the tenth ionization, the atom loses its outermost electrons, and the next electron to be removed is from a much closer shell to the nucleus, which experiences a significantly stronger attraction, thus requiring much more energy

4

Explain how molten magnesium chloride conducts an electric current: Molten magnesium chloride conducts electricity because it dissociates into free-moving ions, specifically

\mathrm{Mg^{2+}}

and

\mathrm{Cl^-}

ions, which can carry charge through the liquid

5

Identify the electrode where oxidation occurs during electrolysis: Oxidation occurs at the anode during the electrolysis of molten magnesium chloride. The half-reaction for oxidation at the anode is:

\mathrm{2Cl^-} \rightarrow \mathrm{Cl_2(g)} + 2e^-

6

Explain why magnesium is not formed during the electrolysis of aqueous magnesium chloride solution: Magnesium is not formed during the electrolysis of aqueous magnesium chloride because water is preferentially reduced over magnesium ions. The half-reaction for the reduction of water is:

\mathrm{2H_2O(l)} + 2e^- \rightarrow \mathrm{H_2(g)} + 2OH^-

Answer

The first ionization energy of magnesium is defined, and the reasons for the trends in ionization energies and electrolysis processes are explained.

Key Concept

Ionization energy trends and electrolysis of magnesium chloride

Explanation

The first ionization energy is the energy needed to remove an electron, and successive ionization energies increase due to stronger nuclear attraction as electrons are removed. The large jump between the tenth and eleventh ionization is due to removing an electron from a new shell. In electrolysis, molten magnesium chloride conducts electricity through free ions, and oxidation occurs at the anode. In aqueous solution, magnesium is not produced due to the preferential reduction of water.

Solution

1

Identify the enthalpy changes: The enthalpy change labeled by

\mathbf{I}

corresponds to the sublimation of magnesium, which requires 148 kJ of energy. The enthalpy change labeled by

\mathbf{V}

corresponds to the lattice enthalpy of magnesium chloride, which is -2326 kJ

2

Calculate lattice enthalpy: To calculate the lattice enthalpy of magnesium chloride, we can use the Born-Haber cycle formula:

\Delta H_{\text{lattice}} = \Delta H_{\text{sublimation}} + \Delta H_{\text{ionization}} + \Delta H_{\text{dissociation}} - \Delta H_{\text{formation}}

Substituting the values:

\Delta H_{\text{lattice}} = 148 \, \text{kJ} + 738 \, \text{kJ} + 1451 \, \text{kJ} - 642 \, \text{kJ} = 1695 \, \text{kJ}

Thus, the calculated lattice enthalpy is +1695 kJ

3

Explain the difference: The difference between the theoretically calculated value (+2326 kJ) and the experimental value (-2326 kJ) can be attributed to factors such as the assumptions made in the Born-Haber cycle, which may not account for all interactions in the solid state, leading to discrepancies in the lattice energy calculations

4

Explain higher lattice enthalpy of magnesium oxide: Magnesium oxide has a higher lattice enthalpy than magnesium chloride due to the smaller ionic radius of oxide ions compared to chloride ions, resulting in stronger electrostatic attractions between the ions in magnesium oxide

5

State the nature of solutions: Aqueous solutions of magnesium oxide are alkaline, while magnesium chloride solutions are neutral

6

Equation for reaction: The equation for the reaction between magnesium oxide and water is:

\text{MgO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Mg(OH)}_2(aq)

Answer

The enthalpy changes are:

\mathbf{I}

= 148 kJ (sublimation of Mg),

\mathbf{V}

= -2326 kJ (lattice enthalpy of MgCl₂). The calculated lattice enthalpy is +1695 kJ, and magnesium oxide has a higher lattice enthalpy due to stronger ionic interactions. Aqueous MgO is alkaline, and the reaction with water produces magnesium hydroxide.

Key Concept

Understanding the Born-Haber cycle and lattice enthalpy calculations is crucial for analyzing ionic compounds.

Explanation

The answers provide insights into the enthalpy changes involved in the formation of magnesium chloride and the comparative analysis of magnesium oxide, highlighting the importance of ionic interactions.

Solution

1

Determine the empirical formula: The alkane contains

82.6\%

by mass of carbon. To find the empirical formula, we assume

100 \, \text{g}

of the alkane, which gives us

82.6 \, \text{g}

of C and

17.4 \, \text{g}

\, \text{of H}

. The moles of C are

\frac{82.6}{12.01} \approx 6.88

and the moles of H are

\frac{17.4}{1.008} \approx 17.28

. The ratio of C to H is approximately

1:2.5

, which simplifies to

2:5

. Thus, the empirical formula is

C_2H_5

2

Deduce the molecular formula: The molar volume of a gas at STP is

22.4 \, \text{L/mol}

. The volume of the gaseous sample is

385 \, \text{cm}^3 = 0.385 \, \text{L}

. The number of moles is

\frac{0.385}{22.4} \approx 0.0172 \, \text{mol}

. The molar mass of the sample is

\frac{1.00 \, \text{g}}{0.0172 \, \text{mol}} \approx 58.14 \, \text{g/mol}

. The empirical formula mass of

C_2H_5

is

2(12.01) + 5(1.008) \approx 29.06 \, \text{g/mol}

. The ratio of molar mass to empirical formula mass is

\frac{58.14}{29.06} \approx 2

, so the molecular formula is

C_4H_{10}

3

State the reagent and conditions for reaction 1: Reaction 1 involves the bromination of the alkane. The reagent is

Br_2

(bromine) and the conditions typically include UV light or heat to initiate the free-radical mechanism

4

Describe the free-radical mechanism: - Initiation:

Br_2 \xrightarrow{UV} 2Br^\cdot

(bromine radicals are formed). - Propagation: -

RCH_3 + Br^\cdot \rightarrow RCH_2Br + H^\cdot

(alkyl radical formation). -

H^\cdot + Br_2 \rightarrow HBr + Br^\cdot

(regenerating bromine radical). - Termination:

Br^\cdot + H^\cdot \rightarrow HBr

(combination of radicals)

5

Explain the SN2 mechanism of reaction 2: In the SN2 mechanism, the nucleophile (

OH^-

) attacks the carbon atom bonded to the leaving group (

Br

) in a single concerted step. The curly arrows show the movement of electron pairs: -

RCH_2Br + OH^- \rightarrow RCH_2OH + Br^-

. The transition state has a pentacoordinate carbon, where the bond to

Br

is breaking while the bond to

OH

is forming

6

Draw the optical isomers: The isomer

X

with the formula

C_4H_9Br

can be represented as two enantiomers. The structures can be drawn with a chiral center, showing the different spatial arrangements of the substituents around the chiral carbon

7

Deduce the rate expression: From the data, we see that when the concentration of

[X]

doubles (from experiment 2 to 3), the rate also doubles. This suggests a first-order dependence on

[X]

. The change in

[OH^-]

does not affect the rate in experiments 1 and 2, indicating zero-order dependence on

[OH^-]

. Thus, the rate expression is:

\text{Rate} = k[X]^1[OH^-]^0 = k[X].

8

Determine the rate constant: Using the data from experiment 3, where the rate is

8.0 \times 10^{-3} \, \text{mol dm}^{-3} \text{min}^{-1}

and

[X] = 4.0 \times 10^{-2} \, \text{mol dm}^{-3}

, we can find

k

:

k = \frac{\text{Rate}}{[X]} = \frac{8.0 \times 10^{-3}}{4.0 \times 10^{-2}} = 0.2 \, \text{min}^{-1}.

9

State the name of isomer X: The isomer

X

is likely 1-bromobutane, as it is a primary alkyl bromide that can undergo hydrolysis with aqueous sodium hydroxide

10

State the equations for the mechanism steps: The slow step is the formation of the alkyl bromide, while the fast step is the nucleophilic attack by

OH^-

. The equations are: - Slow:

RCH_2Br \rightarrow RCH_2^+ + Br^-

. - Fast:

RCH_2^+ + OH^- \rightarrow RCH_2OH

Answer

The empirical formula is

C_2H_5

, the molecular formula is

C_4H_{10}

, and the rate expression is

Rate = k[X]

.

Key Concept

Understanding the empirical and molecular formulas, reaction mechanisms, and rate laws is crucial in organic chemistry.

Explanation

The solution outlines the steps to determine the empirical and molecular formulas, describes the mechanisms of reactions, and deduces the rate expression based on experimental data.

Solution

1

Determine the nature of the reaction: The graph indicates that as the temperature increases, the percentage of ammonia (

\mathrm{NH}_3

) in the equilibrium mixture decreases. This suggests that the forward reaction is exothermic, as increasing temperature shifts the equilibrium to favor the reactants according to Le Chatelier's principle

2

Effect of pressure on yield: Increasing the pressure in the Haber process, which has a decrease in the number of moles of gas from 4 moles of reactants (

\mathrm{N}_2

and

\mathrm{H}_2

) to 2 moles of product (

\mathrm{NH}_3

), will shift the equilibrium to the right, increasing the yield of ammonia

3

Equilibrium constant expression: The equilibrium constant expression for the reaction is given by:

K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3}

4

Calculate $K_c$: Given that the concentration of

\mathrm{NH}_3

at equilibrium is

0.062 \, \mathrm{mol \, dm^{-3}}

, and the initial concentrations of

\mathrm{N}_2

and

\mathrm{H}_2

are

1.00 \, \mathrm{mol \, dm^{-3}}

and

3.00 \, \mathrm{mol \, dm^{-3}}

, respectively, we can substitute these values into the equilibrium expression:

K_c = \frac{(0.062)^2}{(1.00)(3.00)^3} = \frac{0.003844}{27} \approx 0.000142

5

Effect of a catalyst: The presence of a catalyst, such as iron in the Haber process, does not affect the value of the equilibrium constant

K_c

. It only speeds up the rate at which equilibrium is reached

Answer

1. The forward reaction is exothermic.

2. Increasing pressure increases the yield of ammonia.

3.

K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3}

4.

K_c \approx 0.000142

at

400^{\circ}C

.

5. A catalyst does not change

K_c

.

Key Concept

The Haber process is an equilibrium reaction that produces ammonia, and its yield can be influenced by temperature and pressure.

Explanation

The analysis of the graph and the calculations show how temperature and pressure affect the equilibrium position and the equilibrium constant in the Haber process.

Solution

1

Distinguish between strong and weak acids: Strong acids completely dissociate in aqueous solution, while weak acids only partially dissociate. For example, the dissociation of hydrochloric acid (strong acid) can be represented as:

\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}

For hydrocyanic acid (weak acid), the dissociation is represented as:

\mathrm{HCN} \rightleftharpoons \mathrm{H}^{+} + \mathrm{CN}^{-}

2

Deducing the expression for the ionization constant, $K_a$: The ionization constant for hydrocyanic acid is given by the expression:

K_a = \frac{[\mathrm{H}^{+}][\mathrm{CN}^{-}]}{[\mathrm{HCN}]}

To calculate

K_a

from the given

pK_a = 9.21

, we use the formula:

K_a = 10^{-pK_a} = 10^{-9.21} \approx 6.17 \times 10^{-10}

3

Calculate $\left[\mathrm{H}^{+}\right]$ and pH: For a 0.108 mol/dm³ solution of hydrocyanic acid, we assume that the concentration of

\mathrm{H}^{+}

produced is equal to

x

, leading to:

K_a = \frac{x^2}{0.108 - x} \approx \frac{x^2}{0.108}

Solving for

x

gives:

x = \sqrt{K_a \times 0.108} = \sqrt{(6.17 \times 10^{-10}) \times 0.108} \approx 8.06 \times 10^{-6}

Thus,

\left[\mathrm{H}^{+}\right] \approx 8.06 \times 10^{-6}

mol/dm³. The pH is calculated as:

\mathrm{pH} = -\log(8.06 \times 10^{-6}) \approx 5.09

Assumption: The change in concentration of hydrocyanic acid is negligible

Answer

K_a \approx 6.17 \times 10^{-10}

,

\left[\mathrm{H}^{+}\right] \approx 8.06 \times 10^{-6}

mol/dm³,

\mathrm{pH} \approx 5.09

Key Concept

Strong acids fully dissociate, while weak acids partially dissociate in solution.

Explanation

The calculations show how to determine the ionization constant and pH of hydrocyanic acid, illustrating the behavior of weak acids in solution.

Solution

1

Observations to distinguish between acids: When magnesium ribbon is added to nitric acid, effervescence occurs due to the production of hydrogen gas, and the solution may become warm due to the exothermic reaction. In contrast, when added to hydrocyanic acid, the reaction is less vigorous, and there may be no noticeable temperature change

2

Volume calculation for titration: The reaction between nitric acid and sodium hydroxide can be represented as:

\text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}

Using the formula

C_1V_1 = C_2V_2

, where

C_1 = 0.10 \, \text{mol/dm}^3

,

V_1 = 15.0 \, \text{cm}^3

, and

C_2 = 0.20 \, \text{mol/dm}^3

, we can find

V_2

:

V_2 = \frac{C_1V_1}{C_2} = \frac{0.10 \times 15.0}{0.20} = 7.5 \, \text{cm}^3

3

Hypothesis validity: The hypothesis that hydrocyanic acid will react with a smaller volume of sodium hydroxide is not valid. Hydrocyanic acid is a weak acid and will require a larger volume of sodium hydroxide to reach neutralization compared to a strong acid like nitric acid

4

Indicator selection: A suitable indicator for the titration of sodium hydroxide and hydrocyanic acid is phenolphthalein, which changes color in the pH range suitable for detecting the endpoint of a weak acid-strong base titration

5

Identifying Acid 1: Acid 1 is likely nitric acid, as it shows a linear increase in conductivity with concentration, indicating a strong acid that fully dissociates in solution. Acid 2, with its maximum and then decrease in conductivity, suggests it is a weak acid like hydrocyanic acid

Answer

Observations: Effervescence and temperature change for nitric acid; less vigorous reaction for hydrocyanic acid.

Volume of sodium hydroxide required: 7.5 cm³.

Hypothesis: Not valid; hydrocyanic acid requires more volume.

Suitable indicator: Phenolphthalein.

Acid 1 is nitric acid; it shows linear conductivity increase.

Key Concept

Understanding acid-base reactions and conductivity behavior in solutions

Explanation

The observations and calculations help distinguish between strong and weak acids, their reactions with magnesium, and the appropriate titration methods.

Solution

1

Draw the Lewis structure: The Lewis structure for hydrazine,

\\mathrm{N}_{2}\\mathrm{H}_{4}

, shows that each nitrogen atom is bonded to two hydrogen atoms and to each other. The total valence electrons are 14 (5 from two nitrogen atoms and 4 from four hydrogen atoms). The structure is:

H_2N-NH_2

with a single bond between the nitrogen atoms and two single bonds between each nitrogen and its respective hydrogen atoms

2

State and explain the bond angle: The

H-N-H

bond angle in hydrazine is approximately 109.5°, which is characteristic of a tetrahedral arrangement around the nitrogen atoms due to the presence of lone pairs. This angle is influenced by the repulsion between the bonding pairs and the lone pairs of electrons

3

Explain boiling point difference: Hydrazine has a higher boiling point than ethene due to stronger intermolecular forces. Hydrazine exhibits hydrogen bonding due to the presence of N-H bonds, while ethene, being a nonpolar molecule, primarily experiences weaker van der Waals forces

4

Calculate enthalpy change for the reaction: The reaction is:

\\mathrm{N}_{2}\\mathrm{H}_{4}(\\mathrm{l}) + \\mathrm{O}_{2}(\\mathrm{g}) \\rightarrow \\mathrm{N}_{2}(\\mathrm{g}) + 2 \\mathrm{H}_{2}\\mathrm{O}(\\mathrm{l})

Using the enthalpy of formation,

\\Delta H_{f}^{\\ominus}

for products and reactants, we can calculate:

\\Delta H = \\sum \\Delta H_{f}^{\\ominus} (products) - \\sum \\Delta H_{f}^{\\ominus} (reactants)

5

Use bond enthalpy values: To determine the enthalpy change using bond enthalpies, we calculate the total bond energies of the reactants and products. The enthalpy change is given by:

\\Delta H = \\text{Bonds broken} - \\text{Bonds formed}

6

Identify the most accurate calculation: The calculation using the enthalpy of formation is generally more accurate than bond enthalpy calculations because it accounts for the actual formation of the compounds from their elements under standard conditions

7

Calculate $\\Delta S^{\\ominus}$: The change in entropy can be calculated using the standard entropy values:

\\Delta S^{\\ominus} = \\sum S^{\\ominus} (products) - \\sum S^{\\ominus} (reactants)

Substituting the values:

\\Delta S^{\\ominus} = [191 + 2(69.9)] - [121 + 205]

8

Calculate $\\Delta G^{\\ominus}$: The Gibbs free energy change can be calculated using the equation:

\\Delta G^{\\ominus} = \\Delta H^{\\ominus} - T \\Delta S^{\\ominus}

at 298 K

9

Predict spontaneity: The reaction is spontaneous at low temperatures if \\Delta G^{\\ominus} < 0 and at high temperatures if \\Delta S^{\\ominus} > 0, indicating that the reaction favors products as temperature increases. ∻Answer∻ ⚹ The Lewis structure of hydrazine shows a tetrahedral arrangement with a bond angle of approximately 109.5°. Hydrazine has a higher boiling point than ethene due to hydrogen bonding, and the enthalpy change for the reaction can be calculated using formation enthalpies or bond enthalpies. The most accurate method is using formation enthalpies. The entropy change and Gibbs free energy can also be calculated to predict spontaneity at different temperatures. ⚹ ∻Key Concept∻ ⚹ Understanding the molecular structure, intermolecular forces, and thermodynamic properties of hydrazine and its reactions is crucial in predicting behavior and stability. ⚹ ∻Explanation∻ ⚹ The answer provides a comprehensive approach to analyzing hydrazine's properties, including its Lewis structure, boiling point comparison, and thermodynamic calculations, which are essential for understanding its use as a rocket fuel. ⚹

Solution

1

Identify the reaction: The reaction between hydrazine (

\mathrm{N}_{2}\mathrm{H}_{4}

) and hydrochloric acid (

\mathrm{HCl}

) is classified as an acid-base reaction. In this case, hydrazine acts as a base, accepting protons from hydrochloric acid, leading to the formation of the dihydrazinium ion (

\mathrm{N}_{2}\mathrm{H}_{6}^{2+}

)

2

Predict the bond angle: In the dihydrazinium ion (

\mathrm{N}_{2}\mathrm{H}_{6}^{2+}

), the nitrogen atoms are surrounded by four bonding pairs of electrons. This arrangement suggests a tetrahedral geometry, which typically results in a bond angle of approximately 109.5°

3

Determine hybridization: The nitrogen atoms in the dihydrazinium ion (

\mathrm{N}_{2}\mathrm{H}_{6}^{2+}

) exhibit sp³ hybridization. This is due to the presence of four equivalent bonds formed by the mixing of one s orbital and three p orbitals, leading to the tetrahedral arrangement

Answer

The reaction is an acid-base reaction; the bond angle is approximately 109.5°; the hybridization of nitrogen is sp³.

Key Concept

Acid-base reactions involve proton transfer, and hybridization determines molecular geometry and bond angles.

Explanation

Understanding the type of reaction and the hybridization helps predict molecular structure and properties.

It looks like you've shared some biology questions related to cell biology, particularly from the IB Diploma program. While I can help with chemistry concepts, I can also provide some insights into the biology questions you've mentioned. 1. Transport of Monosaccharides: The main method of transport for monosaccharides like fructose across the intestinal epithelium is facilitated diffusion (B). This process involves specific transport proteins that help move the sugars across the cell membrane without using energy. 2. Active Transport in Algae: For the giant marine alga Halicystis ovalis to move sodium ions from vacuoles to seawater through active transport, it requires transmembrane proteins (D). These proteins help in moving ions against their concentration gradient, which is essential for maintaining cellular functions. 3. Cell Membrane Models: The evidence that led to the acceptance of the Singer-Nicolson model over the Davson-Danielli model includes the placement of integral and peripheral proteins in the membrane (C). This model emphasizes the fluid mosaic nature of the membrane, where proteins are embedded within the lipid bilayer. 4. Stem Cells and Stargardt's Disease: The characteristic of stem cells that makes them useful for treating Stargardt's disease is that they can differentiate into retinal cells (A). This ability to become specialized cells is crucial for potential therapies. 5. Striated Muscle Cells and Cell Theory: Striated muscle cells are considered a possible exception to the cell theory because they contain more than one nucleus (B). This multinucleation is atypical for most cells, which usually have a single nucleus. If you have any specific chemistry questions or need further clarification on any of these topics, feel free to ask!

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