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6、在 298 K298 \mathrm{~K} 、定容容器中, 2 mol2 \mathrm{~mol} 固体葈在氧气中燃烧。 已知: $Q_{\mathrm{V}}=...
Mar 15, 2024
6、在 298 K298 \mathrm{~K} 、定容容器中, 2 mol2 \mathrm{~mol} 固体葈在氧气中燃烧。 已知: QV=−10268 kJQ_{\mathrm{V}}=-10268 \mathrm{~kJ}; ΔfHH2O(l)⊖=−285.8 kJ⋅mol−1\Delta_{f} H_{\mathrm{H}_{2} \mathrm{O}(\mathrm{l})}^{\ominus}=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}; ΔfHco2( g)⊖=−393.5 kJ⋅mol−1\Delta_{f} H_{\mathrm{co}_{2}(\mathrm{~g})}^{\ominus}=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}; 试求（1）反应的 ΔrH298⊖\Delta_{r} H_{298}^{\ominus} ；（2）蔡的标准生成热。
Solution
a
Calculate the enthalpy change of the reaction (ΔrH298⊖\Delta_r H_{298}^{\ominus}) using the given heat of reaction at constant volume (QVQ_V) and the stoichiometry of the reaction
b
The standard enthalpy change of the reaction (ΔrH298⊖\Delta_r H_{298}^{\ominus}) is equal to the heat of reaction at constant volume (QVQ_V) for the reaction since the number of moles of gas does not change (solid combustion in oxygen). Thus, ΔrH298⊖=QV=−10268 kJ\Delta_r H_{298}^{\ominus} = Q_V = -10268 \mathrm{~kJ}
c
To find the standard enthalpy of formation of the solid compound (葈), we need to write the balanced chemical equation for its combustion and apply Hess's Law using the standard enthalpies of formation of CO2 and H2O
d
Assume the balanced equation is: 2CxHyOz+(x+y4−z2)O2→2xCO2+yH2O2 \mathrm{C_xH_yO_z} + (x + \frac{y}{4} - \frac{z}{2})O_2 \rightarrow 2xCO_2 + yH_2O
e
Use the standard enthalpies of formation of CO2 and H2O to calculate the standard enthalpy of formation of the solid compound (葈)
f
The standard enthalpy of formation of the solid compound (葈) can be found using the equation: ΔfHCxHyOz(s)⊖=12(ΔrH298⊖−2xΔfHCO2(g)⊖−yΔfHH2O(l)⊖)\Delta_f H_{\mathrm{C_xH_yO_z(s)}}^{\ominus} = \frac{1}{2}(\Delta_r H_{298}^{\ominus} - 2x\Delta_f H_{\mathrm{CO2(g)}}^{\ominus} - y\Delta_f H_{\mathrm{H2O(l)}}^{\ominus})
g
Substitute the known values: ΔfHCxHyOz(s)⊖=12(−10268 kJ−2x(−393.5 kJ/mol)−y(−285.8 kJ/mol))\Delta_f H_{\mathrm{C_xH_yO_z(s)}}^{\ominus} = \frac{1}{2}(-10268 \mathrm{~kJ} - 2x(-393.5 \mathrm{~kJ/mol}) - y(-285.8 \mathrm{~kJ/mol}))
h
Without the values of x, y, and z, we cannot calculate the exact standard enthalpy of formation of the solid compound (葈). Additional information is needed to determine x, y, and z
Answer
The standard enthalpy change of the reaction (ΔrH298⊖\Delta_r H_{298}^{\ominus}) is -10268 kJ. The standard enthalpy of formation of the solid compound (葈) cannot be determined without the values of x, y, and z.
Key Concept
Hess's Law and Standard Enthalpy of Formation: Hess's Law states that the total enthalpy change for a reaction is the same, no matter how many steps or stages the reaction is carried out in. Standard enthalpy of formation is the heat change that results when one mole of a compound is formed from its elements in their standard states.
Explanation
The standard enthalpy change of the reaction was determined directly from the heat of reaction at constant volume. The standard enthalpy of formation of the solid compound requires the balanced chemical equation and the stoichiometric coefficients, which are not provided in this case.
这个固体是萘，重新给出答案
Solution
a
Identify the combustion reaction for naphthalene (C10H8) in oxygen: C10H8(s)+12O2(g)→10CO2(g)+4H2O(l) C_{10}H_{8(s)} + 12O_{2(g)} \rightarrow 10CO_{2(g)} + 4H_{2}O_{(l)}
b
Calculate the standard enthalpy change of the reaction (ΔrH298⊖\Delta_r H_{298}^{\ominus}) using the given heat of reaction at constant volume (QVQ_V) and the stoichiometry of the reaction: ΔrH298⊖=QV=−10268 kJ \Delta_r H_{298}^{\ominus} = Q_V = -10268 \, \text{kJ}
c
Determine the standard enthalpy of formation for naphthalene (ΔfHC10H8(s)⊖\Delta_f H_{\text{C}_{10}\text{H}_{8(s)}}^{\ominus}) using the standard enthalpies of formation for CO2 and H2O: ΔrH298⊖=10ΔfHCO2(g)⊖+4ΔfHH2O(l)⊖−ΔfHC10H8(s)⊖ \Delta_r H_{298}^{\ominus} = 10\Delta_f H_{\text{CO}_{2(g)}}^{\ominus} + 4\Delta_f H_{\text{H}_{2}\text{O}_{(l)}}^{\ominus} - \Delta_f H_{\text{C}_{10}\text{H}_{8(s)}}^{\ominus} −10268 kJ=10(−393.5 kJ/mol)+4(−285.8 kJ/mol)−ΔfHC10H8(s)⊖ -10268 \, \text{kJ} = 10(-393.5 \, \text{kJ/mol}) + 4(-285.8 \, \text{kJ/mol}) - \Delta_f H_{\text{C}_{10}\text{H}_{8(s)}}^{\ominus} −10268 kJ=−3935 kJ−1143.2 kJ−ΔfHC10H8(s)⊖ -10268 \, \text{kJ} = -3935 \, \text{kJ} - 1143.2 \, \text{kJ} - \Delta_f H_{\text{C}_{10}\text{H}_{8(s)}}^{\ominus} ΔfHC10H8(s)⊖=−3935 kJ−1143.2 kJ+10268 kJ \Delta_f H_{\text{C}_{10}\text{H}_{8(s)}}^{\ominus} = -3935 \, \text{kJ} - 1143.2 \, \text{kJ} + 10268 \, \text{kJ} ΔfHC10H8(s)⊖=5189.8 kJ \Delta_f H_{\text{C}_{10}\text{H}_{8(s)}}^{\ominus} = 5189.8 \, \text{kJ}
Answer
The standard enthalpy change of the reaction (ΔrH298⊖\Delta_r H_{298}^{\ominus}) is -10268 kJ. The standard enthalpy of formation for naphthalene (ΔfHC10H8(s)⊖\Delta_f H_{\text{C}_{10}\text{H}_{8(s)}}^{\ominus}) is 5189.8 kJ.
Key Concept
Thermochemistry: The study of the heat energy involved in chemical reactions and changes of physical state.
Explanation
The standard enthalpy change of the reaction was calculated using the given heat of reaction at constant volume, and the standard enthalpy of formation for naphthalene was determined by applying Hess's law and using the standard enthalpies of formation for CO2 and H2O.
6、在 298 K298 \mathrm{~K} 、定容容器中, 2 mol2 \mathrm{~mol} 固体葈在氧气中燃烧。 已知: QV=−10268 kJQ_{\mathrm{V}}=-10268 \mathrm{~kJ}; ΔfHH2O(l)⊖=−285.8 kJ⋅mol−1\Delta_{f} H_{\mathrm{H}_{2} \mathrm{O}(\mathrm{l})}^{\ominus}=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}; ΔfHco2( g)⊖=−393.5 kJ⋅mol−1\Delta_{f} H_{\mathrm{co}_{2}(\mathrm{~g})}^{\ominus}=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}; 试求（1）反应的 ΔrH298⊖\Delta_{r} H_{298}^{\ominus} ；（2）蔡的标准生成热。
Solution
a
Given the heat of combustion for the reaction at constant volume is QV=−10268 kJQ_{\mathrm{V}} = -10268 \mathrm{~kJ} for 2 moles of naphthalene
b
The standard enthalpy change of the reaction (ΔrH298⊖\Delta_{r} H_{298}^{⊖}) is equal to the heat of combustion at constant volume since the process occurs at constant volume. Therefore, ΔrH298⊖=QV=−10268 kJ\Delta_{r} H_{298}^{⊖} = Q_{\mathrm{V}} = -10268 \mathrm{~kJ} for 2 moles of naphthalene
c
To find the standard enthalpy of formation for naphthalene (ΔfHC10H8(s)⊖\Delta_{f} H_{\text{C}_{10}\text{H}_{8(s)}}^{⊖}), we use the balanced chemical equation for the combustion of naphthalene and the standard enthalpies of formation for CO2(g) and H2O(l)
d
The balanced chemical equation for the combustion of naphthalene is: C10H8(s)+12O2(g)→10CO2(g)+4H2O(l)C_{10}H_{8(s)} + 12O_{2(g)} \rightarrow 10CO_{2(g)} + 4H_{2}O(l)
e
Using Hess's law, the standard enthalpy of the reaction can be expressed as: ΔrH298⊖=10ΔfHCO2(g)⊖+4ΔfHH2O(l)⊖−ΔfHC10H8(s)⊖\Delta_{r} H_{298}^{⊖} = 10 \Delta_{f} H_{\mathrm{CO}_{2}(g)}^{⊖} + 4 \Delta_{f} H_{\mathrm{H}_{2}O(l)}^{⊖} - \Delta_{f} H_{\text{C}_{10}\text{H}_{8(s)}}^{⊖}
f
Substituting the given standard enthalpies of formation for CO2(g) and H2O(l) and solving for the standard enthalpy of formation for naphthalene: −10268 kJ=10(−393.5 kJ/mol)+4(−285.8 kJ/mol)−ΔfHC10H8(s)⊖-10268 \mathrm{~kJ} = 10(-393.5 \mathrm{~kJ/mol}) + 4(-285.8 \mathrm{~kJ/mol}) - \Delta_{f} H_{\text{C}_{10}\text{H}_{8(s)}}^{⊖}
g
Solving for ΔfHC10H8(s)⊖\Delta_{f} H_{\text{C}_{10}\text{H}_{8(s)}}^{⊖} gives: ΔfHC10H8(s)⊖=10(−393.5)+4(−285.8)+10268\Delta_{f} H_{\text{C}_{10}\text{H}_{8(s)}}^{⊖} = 10(-393.5) + 4(-285.8) + 10268
h
Calculating the value: ΔfHC10H8(s)⊖=10(−393.5)+4(−285.8)+10268=81.2 kJ/mol\Delta_{f} H_{\text{C}_{10}\text{H}_{8(s)}}^{⊖} = 10(-393.5) + 4(-285.8) + 10268 = 81.2 \mathrm{~kJ/mol} for 1 mole of naphthalene
Answer
(1) The standard enthalpy change of the reaction (ΔrH298⊖\Delta_{r} H_{298}^{⊖}) is -10268 kJ for 2 moles of naphthalene. (2) The standard enthalpy of formation for naphthalene (ΔfHC10H8(s)⊖\Delta_{f} H_{\text{C}_{10}\text{H}_{8(s)}}^{⊖}) is 81.2 kJ/mol.
Key Concept
Hess's Law and Standard Enthalpy of Formation: Hess's Law states that the total enthalpy change for a reaction is the same, no matter how many steps or stages the reaction is carried out in. Standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its elements in their standard states.
Explanation
By using the given heat of combustion and the standard enthalpies of formation for the products, we applied Hess's Law to find the standard enthalpy of formation for naphthalene.
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