ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
6th Edition
ISBN: 9781319306977
Author: LOUDON
Publisher: INTER MAC
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Chapter 4, Problem 4.28P
Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10ΔG°/2.3RT

Where,

ΔG° represents the activation energy for reaction.

R is the gas constant.

T is the temperature.

Expert Solution
Check Mark

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90kJmol1(21.5kcalmol1) and for reaction B is 75kJmol1(17.9kcalmol1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°BΔG°A2.3RT

Where,

ΔG°B and ΔG°A represents the activation energy for reaction A and B.

R is the gas constant.

T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75kJmol190kJmol12.3(8.314JK1mol1)(298K)log(rateArateB)=15kJmol15698.41Jlog(rateArateB)=15×103Jmol15698.41Jmol1log(rateArateB)=2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10ΔG°/2.3RT

Where,

ΔG° represents the activation energy for reaction.

R is the gas constant.

T is the temperature.

Expert Solution
Check Mark

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6K.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90kJmol1(21.5kcalmol1) and for reaction B is 75kJmol1(17.9kcalmol1).

The rate of the reaction is expressed as,

rate=10ΔG°/2.3RT

Where,

ΔG° represents the activation energy for reaction.

R is the gas constant.

T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10ΔG°A/2.3RT'=10ΔG°B/2.3RT

Take log on both sides of equation.

ΔG°A2.3RT'=ΔG°B2.3RT

ΔG°AT'=ΔG°BT …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90kJmol1T'=75kJmol1298KT'=90kJmol1×298K75kJmol1T'=357.6K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6K.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6K.

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