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
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Chapter 18, Problem 18C.4BE
Interpretation Introduction

Interpretation:

Entropy for activation at 450K has to be calculated for the collision between two structureless particles considering M=92gmol-1 and σ=0.45nm2.

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Explanation of Solution

Given information:

    M= 92g/molT= 450Kσ= 0.45nm2

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 = 0.45x10-18m2x(8xkx450K3.14xm2)1/2x6.023x1023mol-1 = 0.45x10-18m2x(2293KxRM)1/2x6.023x1023mol-1 = 2.7103x105m2mol-1x(2293Kx8.314JK-1mol-192x10-3kgmol-1)1/2 = 2.7103x105m2mol-1x(207.21x103Jkg-1)1/2

    1J = 1kgm2s-22.7103x105m2mol-1x(207.21x103Jkg-1)1/2 = 2.7103x105m2mol-1x(207.21x103m2s-2)1/2 = 2.7103x105m2mol-1x(455.203ms-1) = 1233.736x105m3mol-1s-1 = 1.233x108m3mol-1s-1

Activity coefficient is calculated as 1.233x108m3mol-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(450K)26.626x10-34Jsx105Pa = (1.381x10-23JK-1x450K)6.626x10-34Jsx(8.314JK-1mol-1x1kgm2s-21J)x(450K)105Pax1kgm-1s-21Pa = (93.789x1011s-1)(3741.3x10-5m3mol-1) = (93.79x1011s-1)(3.741x10-2m3mol-1) = 350.86x109m3mol-1s-1 = 3.51x1011m3mol-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(ln1.233x108m3mol-1s-13.51x1011m3mol-1s-12) = 8.314JK-1mol-1(9.95) = 82.72JK-1mol-1

The entropy of activation during collision is calculated as 82.72JK-1mol-1.

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Chapter 18 Solutions

Atkins' Physical Chemistry

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