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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 [N 2 O 5 ] T =318K [N 2 O 5 ] 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 E a for tills reaction.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 11, Problem 87AE
Textbook Problem
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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.

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.

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,

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

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