PHYSICAL CHEMISTRY-WEBASSIGN
PHYSICAL CHEMISTRY-WEBASSIGN
11th Edition
ISBN: 9780357087411
Author: ATKINS
Publisher: CENGAGE L
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Chapter 18, Problem 18C.1P
Interpretation Introduction

Interpretation:

Enthalpy, entropy, total energy and Gibbs energy for the given reaction has to be calculated.

Expert Solution & Answer
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Explanation of Solution

Given information:

Frequency factor (A)is given as 4.07×105dm3mol-1s-1.  Activation energy (Ea) is given as 65.4kJmol-1.  Temperature (T) is given as 300K.

Rate constant for the given reaction:

Arrhenius equation for rate constant can be given as,

    kr = A×eEa/RT (1)

Rate constant can be calculated by substituting the values in equation (1) as shown below,

    kr = 4.07×105dm3mol-1s-1×e65.43kJmol-1/(8.314JK-1mol-1×300K) = 4.07×105dm3mol-1s-1×e65.43kJmol-1/(2494.2Jmol-1) = 4.07×105dm3mol-1s-1×e65430Jmol-1/2494.2Jmol-1 = 4.07×105dm3mol-1s-1×e26.232 = 4.07×105dm3mol-1s-1×4.051×1012 = 16.487×107dm3mol-1s-1 = 1.65×106dm3mol-1s-1

Therefore, rate constant is calculated as 1.65×106dm3mol-1s-1.

Enthalpy of Reaction:

The equation for enthalpy of reaction can be given as,

    ΔH = Ea RT (2)

Temperature is given as 300K.

Substituting the values in equation (2), enthalpy of reaction can be obtained as shown below,

    ΔH = 65.43kJmol-1(8.314JK-1mol-1×300K) = (65.43-2.494)kJmol-1 = 62.936kJmol-1

Therefore, enthalpy of reaction is calculated as 62.96kJmol-1.

Gibbs free energy of reaction:

Gibbs free energy of reaction can be given by the equation shown below,

    kr = BeΔGRT (3)

Where,

    T is the temperature.

    R is the gas constant.

    B is coefficient of activation of Gibbs free energy.

B can be calculated as shown below,

    B = kRT2hP0

Where,

    k is Boltzmann constant.

    h is Planck’s constant.

    P0 is pressure of surroundings.

    R is gas constant.

    T is temperature.

    B = 1.381×1023JK-1×8.314JK-1mol-1×(300K)26.626×1034Js×1.00×105Pa = 1.56×1011m3mol-1s-1

Substituting the values in equation (3), Gibbs free energy can be calculated as shown below,

     kr = BeΔGRT1.65×106dm3mol-1s-1 = 1.56×1011m3mol-1s-1×eΔGRT eΔGRT = 1.65×109m3mol-1s-11.56×1011m3mol-1s-1 eΔGRT = 1.06×1020

Taking natural log on both sides,

    ΔGRT = ln(1.06×1020) = 45.99ΔG = 45.99RT = 45.99×8.314JK-1mol-1×300K = 114708.258Jmol-1 = 114.7kJmol-1

Therefore, Gibbs free energy of reaction is 114.7kJmol-1.

Entropy of reaction:

The relationship between Gibbs free energy of reaction, enthalpy of reaction and entropy of reaction can be given by the equation as shown below,

    ΔG = ΔHTΔS

Where,

    ΔS is the entropy of reaction.

Rearranging the equation, the entropy of activation can be obtained as,

    ΔS = ΔHΔGT (4)

Substituting the obtained values in equation (4), the entropy of activation can be calculated.

    ΔS = ΔHΔGT = 62.96kJmol-1114.7kJmol-1300K = 51.74kJmol-1300K = 0.1724kJK-1mol-1 = 172.4JK-1mol-1

Therefore, entropy of reaction is calculated as 172.4JK-1mol-1.

Internal energy of reaction:

Internal energy of the reaction can be given by the equation as shown below,

    ΔU = TΔSPΔV (5)

Where,

    ΔS is the entropy of reaction.

The change in volume is zero as this is a constant volume process.  Hence, the equation (5) becomes as shown below,

    ΔU = TΔS = 300K×172.4JK-1mol-1 = 51720Jmol-1 = 51.72kJmol-1

Therefore, internal energy of reaction is calculated as 51.72kJmol-1.

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Chapter 18 Solutions

PHYSICAL CHEMISTRY-WEBASSIGN

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