# 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

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

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts