Chemistry
Chemistry
13th Edition
ISBN: 9781259911156
Author: Raymond Chang Dr., Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 13, Problem 13.39QP

Some reactions are described as parallel in that the reactant simultaneously forms different products with different rate constants. An example is

A k 2 k 1 B

and

A C

The activation energies are 45.3 kJ/mol for k1 and 69.8 kJ/mol for k2. If the rate constants are equal at 320 K, at what temperature will k1/k2 = 2.00?

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

If the rate constants of given reactions are equal at 320K, then the temperature will k1k2=2.00 has to be identified.

Concept introduction:

Arrhenius equation:

Arrhenius equation is a formula that represents the temperature dependence of reaction rates

The Arrhenius equation can be represented as follows

k=Ae-Ea/RT

Ea is the activation energy ( it is the amount of energy needed to start reaction) and it’s unit is kJ/mol

R represents the universal gas constant and it has the value of 8.314 J/K.mol

T represents the absolute temperature

If  two rate constants at two temperture are known then the activation energy (Ea) can be  calculated using the following equation,

ln(k2k1)=EaR(1T11T2)  Risthegasconstant=8.3145J/molK

Explanation of Solution

Some reactions are described as parallel in that the reactant simultaneously forms different products with different rate constants.  Given example for such type of reactions are,

Ak1BAC

Activation energy for k1is45.3kJ/mol and 69.8kJ/molfork2

If the rate constants of given reactions are equal at 320K, then the temperature will k1k2=2.00 can be determined as follows,

Arrhenius equation is a formula that represents the temperature dependence of reaction rates

The Arrhenius equation can be represented as follows

k=Ae-Ea/RT

The frequency factor A for each of the reaction can be determined as follows using the above equation,

k1=A1eEa/RT=A1e(45300J/mol(8.314J/molK)(320K))=(4.03×108)A1A1=k14.03×108

Similarly frequency factor for reaction k2 can be determined.

k2=A2eEa/RT=A2e(69800J/mol(8.314J/molK)(320K))=(4.04×1012)A1A2=k24.04×1012

Since k1=k2 at 320K, we can write,

A2=k14.04×1012

Assuming, the frequency factor is temperature independent and dividing k1 by k2 to solve for the temperature at which k1/k2=2.00

k1k2=A1eEa1/RTA2eEa2/RT=(k14.03×108k14.04×1012)(eEa1/RTeEa2/RT)

2.00=(1.002×104)(eEa1/RTeEa2/RT)1.996×104=(eEa1/RTeEa2/RT)

Take natural logarithm of both sides of the equation.

ln(1.996×104)=-Ea1RT-(-Ea2RT)=1RT(Ea2-Ea1)9.901=1(8.314J/molK)(T)(69800-45300)J/mol82.32T=24500T=298K

If the rate constants of given reactions are equal at 320K, then at 298K k1k2=2.00.

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Chapter 13 Solutions

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