   # 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. ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

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

## 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=338 K[N2O5] T=338 Kln[N2O5] 0 1.00×10−1 M −2.302 100 6.14×10−2 M −2.79 300 2.33×10−2 M −3.75 600 5.41×10−3 M −5.21 900 1.26×10−3 M −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=318 K[N2O5] T=318 Kln[N2O5] 0 1.00×10−1 M −2.302 100 9.54×10−2 M −2.349 300 8.63×10−2 M −2

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