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The bond lengths of acetylene where found by single crystal neutron diffraction...
Sep 27, 2024
Solution
1
Calculate the mass of the atoms: The mass of hydrogen and carbon in kg can be calculated using their molar masses. For hydrogen: mH=1.0078 g/mol1000=1.0078×10−3 kg/molm_H = \frac{1.0078 \, \text{g/mol}}{1000} = 1.0078 \times 10^{-3} \, \text{kg/mol} and for carbon: mC=12.00 g/mol1000=1.200×10−2 kg/molm_C = \frac{12.00 \, \text{g/mol}}{1000} = 1.200 \times 10^{-2} \, \text{kg/mol}
2
Determine the number of each atom in acetylene: Acetylene (C₂H₂) has 2 carbon atoms and 2 hydrogen atoms. Thus, the total mass of the molecule is: M=2mC+2mH=2(1.200×10−2)+2(1.0078×10−3)M = 2m_C + 2m_H = 2(1.200 \times 10^{-2}) + 2(1.0078 \times 10^{-3})
3
Calculate the moment of inertia: The moment of inertia II for a linear molecule is given by the formula: I=m1r12+m2r22I = m_1 r_1^2 + m_2 r_2^2, where m1m_1 and m2m_2 are the masses of the atoms and r1r_1 and r2r_2 are their distances from the center of mass. The distances for acetylene are: rC=0r_C = 0 (for carbon) and rH=1.188 A˚/2=0.594 A˚=5.94×10−11 mr_H = 1.188 \, \text{Å} / 2 = 0.594 \, \text{Å} = 5.94 \times 10^{-11} \, \text{m}. Thus, I=2mC(0)+2mH(5.94×10−11)2I = 2m_C(0) + 2m_H(5.94 \times 10^{-11})^2
4
Convert the moment of inertia to the required format: After calculating II, convert it to the format I×10−46 kg m2I \times 10^{-46} \, \text{kg m}^2
Answer
[Insert final answer here]
Key Concept
The moment of inertia is a measure of an object's resistance to changes in its rotation, depending on the mass distribution relative to the axis of rotation.
Explanation
The moment of inertia for acetylene is calculated based on the masses of the carbon and hydrogen atoms and their distances from the center of mass, reflecting how the mass is distributed in the molecule.
Solution
1
Calculate the mass of the atoms: The mass of tritium and carbon in kg can be calculated using their molar masses. For tritium: mT=3.01605 g/mol1000=3.01605×10−3 kg/molm_T = \frac{3.01605 \, \text{g/mol}}{1000} = 3.01605 \times 10^{-3} \, \text{kg/mol} and for carbon: mC=12.00 g/mol1000=1.200×10−2 kg/molm_C = \frac{12.00 \, \text{g/mol}}{1000} = 1.200 \times 10^{-2} \, \text{kg/mol}
2
Determine the bond lengths in meters: Convert the bond lengths from angstroms to meters. The C-C bond length is 1.195 A˚=1.195×10−10 m1.195 \, \text{Å} = 1.195 \times 10^{-10} \, \text{m} and the C-T bond length is 1.006 A˚=1.006×10−10 m1.006 \, \text{Å} = 1.006 \times 10^{-10} \, \text{m}
3
Calculate the moment of inertia: The moment of inertia II for tritiated acetylene can be calculated using the formula: I=mC⋅(rCC)2+mT⋅(rCT)2I = m_C \cdot (r_{CC})^2 + m_T \cdot (r_{CT})^2 where rCC=1.195×10−10 mr_{CC} = 1.195 \times 10^{-10} \, \text{m} and rCT=1.006×10−10 mr_{CT} = 1.006 \times 10^{-10} \, \text{m}. Substitute the values: I=(1.200×10−2 kg/mol)⋅(1.195×10−10 m)2+(3.01605×10−3 kg/mol)⋅(1.006×10−10 m)2I = (1.200 \times 10^{-2} \, \text{kg/mol}) \cdot (1.195 \times 10^{-10} \, \text{m})^2 + (3.01605 \times 10^{-3} \, \text{kg/mol}) \cdot (1.006 \times 10^{-10} \, \text{m})^2 Calculate II
4
Convert moment of inertia to rotational constant: The rotational constant BB can be calculated using the formula: B=h8π2cIB = \frac{h}{8 \pi^2 c I} where h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \, \text{Js} and c=299800000 m/sc = 299800000 \, \text{m/s}. Substitute the calculated value of II into this formula to find BB. Finally, convert BB from SI units to cm−1\mathrm{cm}^{-1} by multiplying by 100100 (since 1 m−1=100 cm−11 \, \text{m}^{-1} = 100 \, \text{cm}^{-1}). Round the answer to 4 decimal places
Answer
[Insert final answer here]
Key Concept
The equilibrium rotational constant is a measure of the rotational energy levels of a molecule, which depends on its moment of inertia.
Explanation
The answer is derived from the moment of inertia of the molecule, which is influenced by the masses of the atoms and the bond lengths. The rotational constant is then calculated from this moment of inertia.
Solution
1
Calculate the bond length using the rotational constant: The equilibrium rotational constant Be B_e is related to the bond length r r by the formula Be=h8π2cI B_e = \frac{h}{8 \pi^2 c I} , where I I is the moment of inertia given by I=μr2 I = \mu r^2 . The reduced mass μ \mu for the C-H bond can be calculated as μ=mCmHmC+mH \mu = \frac{m_C m_H}{m_C + m_H} . Using the provided constants, we can find μ \mu and then rearrange the formula to solve for r r
2
Calculate the reduced mass: The mass of hydrogen mH=1.6735×10−27 kg m_H = 1.6735 \times 10^{-27} \, \text{kg} and the mass of carbon mC=1.99265×10−26 kg m_C = 1.99265 \times 10^{-26} \, \text{kg} . Thus, the reduced mass μ \mu is calculated as follows: μ=(1.99265×10−26 kg)(1.6735×10−27 kg)(1.99265×10−26 kg)+(1.6735×10−27 kg)≈1.564×10−27 kg \mu = \frac{(1.99265 \times 10^{-26} \, \text{kg})(1.6735 \times 10^{-27} \, \text{kg})}{(1.99265 \times 10^{-26} \, \text{kg}) + (1.6735 \times 10^{-27} \, \text{kg})} \approx 1.564 \times 10^{-27} \, \text{kg}
3
Calculate the moment of inertia: Using the equilibrium rotational constant for acetylene Be=1.187 cm−1=1.187×100 m−1 B_e = 1.187 \, \text{cm}^{-1} = 1.187 \times 100 \, \text{m}^{-1} , we can rearrange the formula to find r r : r=h8π2cBeμ r = \sqrt{\frac{h}{8 \pi^2 c B_e \mu}} Substituting the values: r=6.626×10−34 Js8π2(299800000 m/s)(1.187×100 m−1)(1.564×10−27 kg)≈1.0634 A˚ r = \sqrt{\frac{6.626 \times 10^{-34} \, \text{Js}}{8 \pi^2 (299800000 \, \text{m/s})(1.187 \times 100 \, \text{m}^{-1})(1.564 \times 10^{-27} \, \text{kg})}} \approx 1.0634 \, \text{Å}
Answer
1.0634
Key Concept
The bond length in a diatomic molecule can be derived from the rotational constant and the reduced mass.
Explanation
The calculated bond length of 1.0634 Å is derived from the relationship between the rotational constant and the moment of inertia, which incorporates the masses of the atoms involved.
Solution
1
Use the Boltzmann distribution: The population of a rotational level J J can be calculated using the Boltzmann equation: NJ=N0⋅(2J+1)e−EJkbTZ N_J = N_0 \cdot \frac{(2J + 1) e^{-\frac{E_J}{k_b T}}}{Z} where N0 N_0 is the population of the ground state, EJ E_J is the energy of the level J J , kb k_b is the Boltzmann constant, T T is the temperature, and Z Z is the partition function. For a rigid rotor, the energy EJ E_J is given by: EJ=hBJ(J+1) E_J = h B J (J + 1)
2
Calculate the energy for \( J = 1 \): Substituting J=1 J = 1 into the energy equation: E1=hB(1)(1+1)=2hB E_1 = h B (1)(1 + 1) = 2hB Using h=6.626×10−34 Js h = 6.626 \times 10^{-34} \, \text{Js} and B=1.1742 cm−1=1.1742×100 m−1 B = 1.1742 \, \text{cm}^{-1} = 1.1742 \times 100 \, \text{m}^{-1} (since 1 cm−1=100 m−1 1 \, \text{cm}^{-1} = 100 \, \text{m}^{-1} ): E1=2⋅(6.626×10−34)⋅(1.1742×100)≈1.554×10−32 J E_1 = 2 \cdot (6.626 \times 10^{-34}) \cdot (1.1742 \times 100) \approx 1.554 \times 10^{-32} \, \text{J}
3
Calculate the partition function \( Z \): For J=0 J = 0 and J=1 J = 1 : Z=∑J=01(2J+1)e−EJkbT=(2⋅0+1)e−E0kbT+(2⋅1+1)e−E1kbT Z = \sum_{J=0}^{1} (2J + 1) e^{-\frac{E_J}{k_b T}} = (2 \cdot 0 + 1) e^{-\frac{E_0}{k_b T}} + (2 \cdot 1 + 1) e^{-\frac{E_1}{k_b T}} Since E0=0 E_0 = 0 : Z=1+3e−1.554×10−32(1.38065×10−23)(257) Z = 1 + 3 e^{-\frac{1.554 \times 10^{-32}}{(1.38065 \times 10^{-23})(257)}} Calculating the exponent: −1.554×10−32(1.38065×10−23)(257)≈−0.000046 -\frac{1.554 \times 10^{-32}}{(1.38065 \times 10^{-23})(257)} \approx -0.000046 Thus, Z≈1+3e−0.000046≈1+3⋅0.999954≈3.9998 Z \approx 1 + 3 e^{-0.000046} \approx 1 + 3 \cdot 0.999954 \approx 3.9998
4
Calculate the population \( N_1 \): Now substituting back into the Boltzmann equation: N1=1000⋅(2⋅1+1)e−E1kbTZ=1000⋅3e−1.554×10−32(1.38065×10−23)(257)3.9998 N_1 = 1000 \cdot \frac{(2 \cdot 1 + 1) e^{-\frac{E_1}{k_b T}}}{Z} = 1000 \cdot \frac{3 e^{-\frac{1.554 \times 10^{-32}}{(1.38065 \times 10^{-23})(257)}}}{3.9998} Calculating N1 N_1 : N1≈1000⋅3⋅0.9999543.9998≈749.9 N_1 \approx 1000 \cdot \frac{3 \cdot 0.999954}{3.9998} \approx 749.9 ∻Answer∻ ⚹ 749.9 ⚹ ∻Key Concept∻ ⚹ The population of rotational levels in a molecule follows a Boltzmann distribution, which is influenced by temperature and energy levels. ⚹ ∻Explanation∻ ⚹ The calculated population of the rotational level J=1 J=1 at 257 K is approximately 749.9 molecules, based on the Boltzmann distribution and the given constants. ⚹
Solution
1
Calculate the reduced mass: The reduced mass μ\mu for the vibrational frequency can be calculated using the formula μ=m1m2m1+m2\mu = \frac{m_1 m_2}{m_1 + m_2}. For 1H1H{ }^{1} \mathrm{H}^{1} \mathrm{H} (two hydrogen atoms), the mass is mH=1.0078 g/mol1000=1.0078×10−3 kg/molm_H = \frac{1.0078 \, \text{g/mol}}{1000} = 1.0078 \times 10^{-3} \, \text{kg/mol}. Thus, μH=(1.0078×10−3)22×1.0078×10−3=5.039×10−4 kg\mu_H = \frac{(1.0078 \times 10^{-3})^2}{2 \times 1.0078 \times 10^{-3}} = 5.039 \times 10^{-4} \, \text{kg}. For 3H3H{ }^{3} \mathrm{H}^{3} \mathrm{H} (two tritium atoms), the mass is mT=3.01605 g/mol1000=3.01605×10−3 kg/molm_T = \frac{3.01605 \, \text{g/mol}}{1000} = 3.01605 \times 10^{-3} \, \text{kg/mol}. Thus, μT=(3.01605×10−3)22×3.01605×10−3=1.508×10−3 kg\mu_T = \frac{(3.01605 \times 10^{-3})^2}{2 \times 3.01605 \times 10^{-3}} = 1.508 \times 10^{-3} \, \text{kg}
2
Calculate the vibrational frequency: The vibrational frequency ν\nu can be calculated using the formula ν=12πkμ\nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}. For 1H1H{ }^{1} \mathrm{H}^{1} \mathrm{H}, νH=12π576.0 N/m5.039×10−4 kg≈11.56 Hz\nu_H = \frac{1}{2\pi} \sqrt{\frac{576.0 \, \text{N/m}}{5.039 \times 10^{-4} \, \text{kg}}} \approx 11.56 \, \text{Hz}. For 3H3H{ }^{3} \mathrm{H}^{3} \mathrm{H}, νT=12π576.0 N/m1.508×10−3 kg≈7.99 Hz\nu_T = \frac{1}{2\pi} \sqrt{\frac{576.0 \, \text{N/m}}{1.508 \times 10^{-3} \, \text{kg}}} \approx 7.99 \, \text{Hz}
3
Calculate the frequency difference: The absolute value of the frequency difference is ∣νH−νT∣=∣11.56−7.99∣≈3.57 Hz|\nu_H - \nu_T| = |11.56 - 7.99| \approx 3.57 \, \text{Hz}. To convert this to cm−1\mathrm{cm}^{-1}, use the conversion 1 Hz=1c cm−11 \, \text{Hz} = \frac{1}{c} \, \text{cm}^{-1}, where c=299800000 m/sc = 299800000 \, \text{m/s}. Thus, the difference in cm−1\mathrm{cm}^{-1} is 3.57299800000≈1.19×10−8 cm−1\frac{3.57}{299800000} \approx 1.19 \times 10^{-8} \, \mathrm{cm}^{-1}. Rounding to 1 cm−11 \, \mathrm{cm}^{-1} gives the final answer
Answer
1
Key Concept
The vibrational frequency difference between isotopes is calculated using reduced mass and Hooke's law.
Explanation
The difference in vibrational frequencies arises from the change in reduced mass when substituting hydrogen with tritium, affecting the vibrational motion of the molecule.
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