Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 18, Problem 18C.4AE
Interpretation Introduction

Interpretation:

Entropy for activation at 300K has to be calculated for the collision between two structureless particles considering M=65gmol-1 and σ=0.35nm2.

Expert Solution & Answer
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Explanation of Solution

Given information:

    M= 65g/molT= 300Kσ= 0.35nm2

Entropy of collisions between two particles can be calculated using the formula,

    ΔS = R(lnAB2)(1)

Where,

    A is the activity coefficient.

    ΔS is the entropy of collision.

    R is the gas constant.

Activity coefficient can be expressed as given below,

  A = σ(8kTπμ)1/2NA(2)

Where,

    μ is reduced mass

    σ is collision cross section area

    NA is Avogadro number

    k is the Boltzmann constant

Calculation for identification of activity coefficient:

As the collision is between two identical particles, the reduced mass is m/2.

Boltzmann constant and gas constant is related as shown below,

    km = RM

Substituting the values in equation (2), activity coefficient can be obtained as shown below,

    A = σ(8kTπμ)1/2NA = 0.35x10-18m2x(8xkx300K3.14xm2)1/2x6.023x1023mol-1 = 0.35x10-18m2x(1528KxRM)1/2x6.023x1023mol-1 = 2.10805x105m2mol-1x(1528Kx8.314JK-1mol-165x10-3kgmol-1)1/2 = 2.10805x105m2mol-1x(195.44x103Jkg-1)1/2

    1J = 1kgm2s-22.10805x105m2mol-1x(195.44x103Jkg-1)1/2 = 2.10805x105m2mol-1x(195.44x103m2s-2)1/2 = 2.10805x105m2mol-1x(442.085ms-1) = 931.9505x105m3mol-1s-1 = 0.93x108m3mol-1s-1

Activity coefficient is calculated as 0.93x108m3mol-1s-1.

Entropy of activation:

The rate constant equation can also be given as shown below by use of Eyring equation.

    kr = BeΔS/ReΔH/RTWhere, B = (kTh)(RTpO) = kRT2hpO

B can be calculated as shown below,

B = kRT2hpO = (1.381x10-23JK-1)x(8.314JK-1mol-1)x(300K)26.626x10-34Jsx105Pa = (1.381x10-23JK-1x300K)6.626x10-34Jsx(8.314JK-1mol-1x1kgm2s-21J)x(300K)105Pax1kgm-1s-21Pa = (62.52x1011s-1)(2494.2x10-5m3mol-1) = (62.52x1011s-1)(2.494x10-2m3mol-1) = 155.92x109m3mol-1s-1 = 1.56x1011m3mol-1s-1

Substitution of the obtained values in the equation (1), entropy of activation can be calculated as shown below,

  ΔS = R(lnAB2) = 8.314JK-1mol-1(ln0.93x108m3mol-1s-11.56x1011m3mol-1s-12) = 8.314JK-1mol-1(9.425) = 78.35JK-1mol-1

The entropy of activation during collision is calculated as 78.35JK-1mol-1.

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Chapter 18 Solutions

Atkins' Physical Chemistry

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