   Chapter 12, Problem 72AE

Chapter
Section
Textbook Problem

# Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to 35°C?

Interpretation Introduction

Interpretation: It is assumed by chemist that, an increase of 10K temperature doubles the rate of reaction. The activation energy is calculated for the given temperature to make the given statement correct.

Concept introduction: The energy difference between activated complex and reactants is known as activation energy. The relationship between the rate constant and temperature is given by the Arrhenius equation,

k=AeEaRT

To determine: The value of activation energy to make the statement “an increase of 10K temperature doubles the rate of reaction” true for a temperature increases from 25°C to 35°C .

Explanation

Given

The initial temperature is 25°C .

The final temperature is 35°C .

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence, the conversion of 25°C into Kelvin is,

T(K)=T(°C)+273T(K)=(25+273)K=298K

The conversion of 35°C into Kelvin is,

T(K)=T(°C)+273T(K)=(35+273)K=308K

It is assumed that the rate constant at 298K is k1=k . Since, as the temperature is increased from 298K to 308K , the rate constant is also increased by a factor of 2 . Therefore, the rate constant at higher temperature is,

k2=2k

Formula

The activation energy is calculated by using the formula,

ln(k2k1)=EaR(1T11T2)

Where,

• k1 is the rate constant at initial temperature.
• k2 is the rate constant at final temperature.
• T1 isinitial temperature.
• T2 isfinal temperature.
• Ea is the activation energy.
• R is the universal gas constant (8.314J/Kmol) .

Substitute the values of R,k1,k2,T1 and T2 in the above expression

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