Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 12, Problem 91AE

For the reaction

2N 2 O 5 ( g ) 4 NO 2 ( g ) + O 2 ( g )

the following data were collected, where

Rate Δ [ N 2 O 5 ] Δ t

Time (s) T = 338 K [N2O5]

T=318K

[N2O5]

0 1.00 × 10−1 M 1.00 × 10−1 M
100. 6.14 × 10−2 M 9.54 × 10−2 M
300 2.33 × 10−2 M 8.63 × 10−2 M
600. 5.41 × 1O−3 M 7.43 × 10−2 M
900. 1.26 × 1O−3 M 6.39 × 10−2 M

Calculate Ea for tills reaction.

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation: The decomposition reaction of (N2O5) , its rate law and concentration of (N2O5) at different temperature and time is given. The value of Ea is to be calculated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

t12=0.693k

To determine: The value of Ea for the given reaction.

Answer to Problem 91AE

Answer

The value of Ea for the given reaction is 1.01×105J/mol_ .

Explanation of Solution

Explanation

The reaction that takes place is,

2N2O5(g)4NO2(g)+O2(g)

The value of ln(N2O5) at 338K is calculated using the given table. That is,

Time (s) T=338K[N2O5] T=338Kln[N2O5]
0 1.00×101M 2.302
100 6.14×102M 2.79
300 2.33×102M 3.75
600 5.41×103M 5.21
900 1.26×103M 6.67

The graph is plotted for ln[N2O5] versus time at T=338K is,

Chemistry, Chapter 12, Problem 91AE , additional homework tip  1

Since, the obtained graph is a straight line, hence, the order of the reaction is first. The slope of graph is 0.0048 .

The integral rate law equation of first order reaction is,

ln[N2O2]=k1t+ln[N2O2]0 (1)

Where,

  • k1 is the rate constant at 338K .
  • [N2O2]0 is the initial concentration of reactant.
  • t is the time.

The equation (1) is similar to the equation of a straight line, that is,

y=mx+c (2)

Where,

  • y is the y-intercept.
  • x is the x-intercept.
  • m is the slope.
  • c is a constant.

Compare equation (1) and (2).

m=k1

Substitute the value of slope in the above equation.

m=k1k1=(0.0048s1)=0.0048s1_

The value of rate constant at T=318K is 4.977×104s1_ .

The value of ln(N2O5) at 318K is calculated using the given table. That is,

Time (s) T=318K[N2O5] T=318Kln[N2O5]
0 1.00×101M 2.302
100 9.54×102M 2.349
300 8.63×102M 2.449
600 7.43×102M 2.599
900 6.39×102M 2.750

The graph is plotted for ln[N2O5] versus time at T=318K is,

Chemistry, Chapter 12, Problem 91AE , additional homework tip  2

Since, the obtained graph is a straight line, hence, the order of the reaction is first. The slope of graph is 0.0004977 .

The integral rate law equation of first order reaction is,

ln[N2O2]=k2t+ln[N2O2]0 (3)

Where,

  • k2 is the rate constant at 318K .
  • [N2O2]0 is the initial concentration of reactant.
  • t is the time.

The equation (3) is similar to the equation of a straight line, that is,

y=mx+c (4)

Where,

  • y is the y-intercept.
  • x is the x-intercept.
  • m is the slope.
  • c is a constant.

Compare equation (3) and (4).

m=k2

Substitute the value of slope in the above equation.

m=k2k2=(0.0004977s1)=4.977×104s1_

The value of Ea for the given reaction is 1.01×105J/mol_ .

The value of rate constant at T=338K is 0.0048s1 .

The value of rate constant at T=318K is 4.977×104s1 .

Formula

The Arrhenius equation is,

lnk2k1=EaR(1T21T1)

Where,

  • Ea is the activation energy.
  • R is the universal gas constant (8.314J/Kmol) .
  • T1 is the absolute temperature (338K) .
  • T2 is the absolute temperature (318K) .

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

lnk2k1=EaR(1T21T1)ln(4.977×104s10.048s1)=Ea(8.314J/Kmol)(1(318K)1(338K))Ea=1.01×105J/mol_

Conclusion

Conclusion

The calculated value of Ea for the given reaction is 1.01×105J/mol_

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