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If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).
[Given: Gm - Gm(0) is _____kJmol-1.
[Given: Gm - Gm(0) is _____kJmol-1.
Correct answer is '-77.18'. Can you explain this answer?
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If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 1...
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If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 1...
The change in rotational free energy (ΔG_rot) can be calculated using the formula:
ΔG_rot = -RTln(Q_rot)
where R is the gas constant, T is the temperature in Kelvin, and Q_rot is the partition function for rotational motion.
To calculate Q_rot, we need to use the rotational constants A, B, and C for H2O(g) at 15000K.
The partition function for rotational motion can be calculated as follows:
Q_rot = (8π^2IkT) / (h^2cA) × (8π^2IkT) / (h^2cB) × (8π^2IkT) / (h^2cC)
where I is the moment of inertia of the molecule, k is Boltzmann's constant, h is Planck's constant, and c is the speed of light.
To calculate the moment of inertia, we can use the formula:
I = μr^2
where μ is the reduced mass of the molecule and r is the bond length.
For H2O, the bond length (r) is approximately 0.96 Å and the reduced mass (μ) can be calculated as:
μ = (mH * mO) / (mH + mO)
where mH is the mass of hydrogen and mO is the mass of oxygen.
Substituting the values into the equation, we can calculate the reduced mass:
μ = (2 * 16.00 amu) / (2 + 16.00 amu) = 2.67 amu
Next, we can calculate the partition function:
Q_rot = (8π^2 * 2.67 amu * 1.38 J/K * 15000 K) / (6.63 x 10^-34 J·s)^2 * (2.998 x 10^8 m/s) * 27.8778 cm^-1) × (8π^2 * 2.67 amu * 1.38 J/K * 15000 K) / (6.63 x 10^-34 J·s)^2 * (2.998 x 10^8 m/s) * 14.5042 cm^-1) × (8π^2 * 2.67 amu * 1.38 J/K * 15000 K) / (6.63 x 10^-34 J·s)^2 * (2.998 x 10^8 m/s) * 9.286 cm^-1)
Calculating the value, we get:
Q_rot ≈ 1.59 x 10^49
Finally, we can substitute the values into the equation for ΔG_rot:
ΔG_rot = -RTln(Q_rot)
Substituting the values:
ΔG_rot = -(8.314 J/(mol·K))(15000 K) ln(1.59 x 10^49)
Calculating the natural logarithm and the negative sign, we get:
ΔG_rot ≈ -(-77.18 kJ/mol) ≈ 77.18 kJ/mol
Therefore, the value of ΔG_rot is approximately -77.18 kJ/mol.
ΔG_rot = -RTln(Q_rot)
where R is the gas constant, T is the temperature in Kelvin, and Q_rot is the partition function for rotational motion.
To calculate Q_rot, we need to use the rotational constants A, B, and C for H2O(g) at 15000K.
The partition function for rotational motion can be calculated as follows:
Q_rot = (8π^2IkT) / (h^2cA) × (8π^2IkT) / (h^2cB) × (8π^2IkT) / (h^2cC)
where I is the moment of inertia of the molecule, k is Boltzmann's constant, h is Planck's constant, and c is the speed of light.
To calculate the moment of inertia, we can use the formula:
I = μr^2
where μ is the reduced mass of the molecule and r is the bond length.
For H2O, the bond length (r) is approximately 0.96 Å and the reduced mass (μ) can be calculated as:
μ = (mH * mO) / (mH + mO)
where mH is the mass of hydrogen and mO is the mass of oxygen.
Substituting the values into the equation, we can calculate the reduced mass:
μ = (2 * 16.00 amu) / (2 + 16.00 amu) = 2.67 amu
Next, we can calculate the partition function:
Q_rot = (8π^2 * 2.67 amu * 1.38 J/K * 15000 K) / (6.63 x 10^-34 J·s)^2 * (2.998 x 10^8 m/s) * 27.8778 cm^-1) × (8π^2 * 2.67 amu * 1.38 J/K * 15000 K) / (6.63 x 10^-34 J·s)^2 * (2.998 x 10^8 m/s) * 14.5042 cm^-1) × (8π^2 * 2.67 amu * 1.38 J/K * 15000 K) / (6.63 x 10^-34 J·s)^2 * (2.998 x 10^8 m/s) * 9.286 cm^-1)
Calculating the value, we get:
Q_rot ≈ 1.59 x 10^49
Finally, we can substitute the values into the equation for ΔG_rot:
ΔG_rot = -RTln(Q_rot)
Substituting the values:
ΔG_rot = -(8.314 J/(mol·K))(15000 K) ln(1.59 x 10^49)
Calculating the natural logarithm and the negative sign, we get:
ΔG_rot ≈ -(-77.18 kJ/mol) ≈ 77.18 kJ/mol
Therefore, the value of ΔG_rot is approximately -77.18 kJ/mol.
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If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).[Given: Gm - Gm(0) is _____kJmol-1.Correct answer is '-77.18'. Can you explain this answer?
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If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).[Given: Gm - Gm(0) is _____kJmol-1.Correct answer is '-77.18'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).[Given: Gm - Gm(0) is _____kJmol-1.Correct answer is '-77.18'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).[Given: Gm - Gm(0) is _____kJmol-1.Correct answer is '-77.18'. Can you explain this answer?.
If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).[Given: Gm - Gm(0) is _____kJmol-1.Correct answer is '-77.18'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).[Given: Gm - Gm(0) is _____kJmol-1.Correct answer is '-77.18'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If rotational constant for H2O(g) at 15000K are A = 27.8778cm-1, B = 14.5042cm-1 and C = 9.286cm-1. The value of change in rotational free energy. (Round off to two decimal places).[Given: Gm - Gm(0) is _____kJmol-1.Correct answer is '-77.18'. Can you explain this answer?.
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