ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 13, Problem 13F.5P
Interpretation Introduction

Interpretation:

The value of GmΘ(200K)GmΘ(0) for Cl2O2 has to be calculated.

Concept introduction:

A thermodynamic potential that is utilized in the calculation of the highest reversible function taking place in the thermodynamic system at the constant value pressure and temperature is known as Gibbs-free energy.  It is denoted by ΔG°.

Expert Solution & Answer
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Answer to Problem 13F.5P

The value of GmΘ(200K)GmΘ(0) for Cl2O2 is 4.576×104Jmol-1_.

Explanation of Solution

The given rotational constants for Cl2O2 are B=13109.4,2409.8 and 2139.7MHz.

The given rotational symmetry number is 2.

The given vibtational wavenumbers for Cl2O2 are 753,542,310,127,646 and 419cm1 which are all degenerate.

The molar mass of Cl2O2 is 102.9g/mol.

The given temperature is 200K.

The value of Planck’s constant is 6.626×1034Js.

The value of speed of light is 2.998×1010cm/s.

The expression that is used to calculate the standard molar Gibbs energy, GmΘ, is mentioned as follows.

    GmΘ(2000K)GmΘ(0)=RTlnqm0NA        (1)

Where,

  • qm0 is the molecular partition function.
  • R is the universal gas constant.
  • T is the temperature.

In the above expression, qm0NA is equal to qmTNA=Translationalpartitionfunction, qmRNA=Rotationalpartitionfunction and qmVNA=Vibrationalpartitionfunction.

The expression for translational partition function, qmTNA for Cl2O2 is given below.

    qmTNA=kTpΘ(2πmkTh)3=(2πk)3/2pΘh3(T/K)5/2(M/gmol1)3/2

Where,

  • h is the Planck’s constant.
  • M is the molar mass of Cl2O2.
  • k is the Boltzmann constant.(1.38×1023JK1)

Substitute (2πk)3/2pΘh3 as 2.561×102, T as 200K and M as 102.9g/mol in the above equation.

    qmTNA=2.561×102×(200)5/2(102.9)3/2=2.561×102×(565685.425)×(1043.8139)=1.512×107

Thus the value of translational partition function for Cl2O2 is 1.512×107.

The expression for rotational partition function, qmRNA for Cl2O2 which is a non linear molecule given below.

    qmR=1σkThc(πABC)1/2        (2)

Where,

  • k is the Boltzmann constant.(1.38×1023JK1)
  • h is the Planck’s constant.
  • c is the speed of light.
  • ABC are the rotational constants of the molecules along three directions.
  • σ is the symmetry number equals to 2.

Substitute the values of k, h, c and σ in the above equation.

    qmR=12((1.38×1023JK1)T6.626×1034J s×2.998×1010cm/s)1.5(3.14ABC)0.5=12((1.38×1023K1)1.5(3.14)0.5(6.626×1034J s×2.998×1010cm/s)1.5)((T)1.5(ABC)0.5)=1.02702((T/K)1.5(c)1.5(ABC/cm3)0.5)        (3)

Substitute the value of T=200K and c as 2.998×1010cm/s, A as 13109.4MHz B as 2409.8MHz and C as 2139.7MHz in the above equation.

    qmR=1.02702((200)1.5(2.998×1010cm/s)1.5((13109.4×2409.8×2139.7)106/sMHz/cm3)0.5)=1.02702(1.46822×1019((6.7595×1010)106/sMHz)0.5)=2.9×104

Thus, the value of rotational partition function of Cl2O2 is 2.9×104.

As, Cl2O2 possesses six vibrational degrees of freedom and all are having partition function, so, the value of vibrational partition function for the vibrational degrees of freedom of Cl2O2  is given below.

    qmV=(11e1.4388(νcm1¯)T/K)        (4)

Where,

  • ν¯ is the vibrational wave number.

The six vibrational modes of Cl2O2 are labeled as (1), (2), (3), (4), (5) and (6).

Substitute the value of ν¯ as 753cm1 and temperature in equation (4).

    qmV(1)=(11e1.4388(753cm1)200)=1.004

Substitute the value of ν¯ as 542cm1 and temperature in equation (4).

    qmV(2)=(11e1.4388(542cm1)200)=1.021

Substitute the value of ν¯ as 310cm1 and temperature in equation (4).

    qmV(3)=(11e1.4388(310cm1)200)=1.120

Substitute the value of ν¯ as 127cm1 and temperature in equation (4).

    qmV(4)=(11e1.4388(127cm1)200)=1.670

Substitute the value of ν¯ as 646cm1 and temperature in equation (4).

    qmV(5)=(11e1.4388(646cm1)200)=1.010

Substitute the value of ν¯ as 419cm1 and temperature in equation (4).

    qmV(6)=(11e1.4388(419cm1)200)=1.052

So, the total vibrational partition function is calculated below.

    qmV=i=16qmV(i)=qmV(1)×qmV(2)×qmV(3)×qmV(4)×qmV(5)×qmV(6)

Substitute the value of qmV(1)×qmV(2)×qmV(3)×qmV(4)×qmV(5)×qmV(6) in the above equation.

    qmV=1.004×1.021×1.120×1.670×1.010×1.052=2.037

Substitute the value of qmE, qmT, qmR, R as 8.3145Jmol1K1, T and qmV for Cl2O2 in equation (1).

    GmΘ(200K)GmΘ(0)=8.3145Jmol1K1×200Kln[(1.512×107)(2.9×104)(2.037)(1)]=1662.9×ln(8.9318×1011)=4.576×104Jmol-1_

Therefore, the value of GmΘ(200K)GmΘ(0) for Cl2O2 is 4.576×104Jmol-1_.

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Chapter 13 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 13 - Prob. 13A.1AECh. 13 - Prob. 13A.1BECh. 13 - Prob. 13A.2AECh. 13 - Prob. 13A.2BECh. 13 - Prob. 13A.3AECh. 13 - Prob. 13A.3BECh. 13 - Prob. 13A.4AECh. 13 - Prob. 13A.4BECh. 13 - Prob. 13A.5AECh. 13 - Prob. 13A.5BECh. 13 - Prob. 13A.6AECh. 13 - Prob. 13A.6BECh. 13 - Prob. 13A.1PCh. 13 - Prob. 13A.2PCh. 13 - Prob. 13A.4PCh. 13 - Prob. 13A.5PCh. 13 - Prob. 13A.6PCh. 13 - Prob. 13A.7PCh. 13 - Prob. 13B.1DQCh. 13 - Prob. 13B.2DQCh. 13 - Prob. 13B.3DQCh. 13 - Prob. 13B.1AECh. 13 - Prob. 13B.1BECh. 13 - Prob. 13B.2AECh. 13 - Prob. 13B.2BECh. 13 - Prob. 13B.3AECh. 13 - Prob. 13B.3BECh. 13 - Prob. 13B.4AECh. 13 - Prob. 13B.4BECh. 13 - Prob. 13B.7AECh. 13 - Prob. 13B.7BECh. 13 - Prob. 13B.8AECh. 13 - Prob. 13B.8BECh. 13 - Prob. 13B.9AECh. 13 - Prob. 13B.9BECh. 13 - Prob. 13B.10AECh. 13 - Prob. 13B.10BECh. 13 - Prob. 13B.11AECh. 13 - Prob. 13B.11BECh. 13 - Prob. 13B.12AECh. 13 - Prob. 13B.12BECh. 13 - Prob. 13B.4PCh. 13 - Prob. 13B.5PCh. 13 - Prob. 13B.6PCh. 13 - Prob. 13B.7PCh. 13 - Prob. 13B.8PCh. 13 - Prob. 13B.10PCh. 13 - Prob. 13C.1DQCh. 13 - Prob. 13C.2DQCh. 13 - Prob. 13C.1AECh. 13 - Prob. 13C.1BECh. 13 - Prob. 13C.6AECh. 13 - Prob. 13C.6BECh. 13 - Prob. 13C.7AECh. 13 - Prob. 13C.7BECh. 13 - Prob. 13C.3PCh. 13 - Prob. 13C.7PCh. 13 - Prob. 13C.8PCh. 13 - Prob. 13C.9PCh. 13 - Prob. 13D.1DQCh. 13 - Prob. 13D.2DQCh. 13 - Prob. 13D.3DQCh. 13 - Prob. 13D.4DQCh. 13 - Prob. 13D.1AECh. 13 - Prob. 13D.1BECh. 13 - Prob. 13D.1PCh. 13 - Prob. 13D.2PCh. 13 - Prob. 13E.1DQCh. 13 - Prob. 13E.2DQCh. 13 - Prob. 13E.3DQCh. 13 - Prob. 13E.4DQCh. 13 - Prob. 13E.5DQCh. 13 - Prob. 13E.6DQCh. 13 - Prob. 13E.1AECh. 13 - Prob. 13E.1BECh. 13 - Prob. 13E.2AECh. 13 - Prob. 13E.2BECh. 13 - Prob. 13E.3AECh. 13 - Prob. 13E.3BECh. 13 - Prob. 13E.4AECh. 13 - Prob. 13E.4BECh. 13 - Prob. 13E.5AECh. 13 - Prob. 13E.5BECh. 13 - Prob. 13E.6AECh. 13 - Prob. 13E.6BECh. 13 - Prob. 13E.7AECh. 13 - Prob. 13E.7BECh. 13 - Prob. 13E.8AECh. 13 - Prob. 13E.8BECh. 13 - Prob. 13E.9AECh. 13 - Prob. 13E.9BECh. 13 - Prob. 13E.1PCh. 13 - Prob. 13E.2PCh. 13 - Prob. 13E.3PCh. 13 - Prob. 13E.4PCh. 13 - Prob. 13E.7PCh. 13 - Prob. 13E.9PCh. 13 - Prob. 13E.10PCh. 13 - Prob. 13E.11PCh. 13 - Prob. 13E.14PCh. 13 - Prob. 13E.15PCh. 13 - Prob. 13E.16PCh. 13 - Prob. 13E.17PCh. 13 - Prob. 13F.1DQCh. 13 - Prob. 13F.2DQCh. 13 - Prob. 13F.3DQCh. 13 - Prob. 13F.1AECh. 13 - Prob. 13F.1BECh. 13 - Prob. 13F.2AECh. 13 - Prob. 13F.2BECh. 13 - Prob. 13F.3AECh. 13 - Prob. 13F.3BECh. 13 - Prob. 13F.3PCh. 13 - Prob. 13F.4PCh. 13 - Prob. 13F.5PCh. 13 - Prob. 13F.6PCh. 13 - Prob. 13.1IA
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