Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
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Chapter 13, Problem 13.69QA
Interpretation Introduction

To find:

a) In the first order reaction of decomposition of N2O5  whatamount (concentration) of N2O5  remains in the solution after 1 h if the initial concentration was 0.50 mol/ L and k=6.32×10-4s-1?

b) Percentage of N2O5 reacted after 1 h.

Expert Solution & Answer
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Answer to Problem 13.69QA

Solution:

a) The amount (concentration) of N2O5  remains in the solution after 1 h is,0.051 mol/L.

b) Percent of N2O5 reacted = 90 %.

Explanation of Solution

1) Concept:

In the first order reaction, the rate nearly depends on only one reactant, so it is possible to find out the concentration of the reactant at any time once the reaction gets started using the integrated rate law.

2) Formula:

i. Integrated rate law for first order reaction is

ln[X]t[X]0= -kt …(equation 1)

i.e. ln [X]t= -kt+ln [X]0

Where

[X]t= concentration of reactant at time t

[X]0= initial concentration of reactant

k= rate constant

t= reaction time

ii. Percent reacted:

Reacted concentration = Initial concentration  Unreacted concentration

Percent reacted= Reacted cocentrationInitial cocentration×100

3) Given:

[X]0=0.50 mol/L

k=6.32×10-4s-1

t=1 h

[X]t= ?

4) Calculation:

a) Amount (concentration) of N2O5  remains in the solution after 1 h:

Since rate constant is in second (s) unit, we need to convert time (t) into seconds (s).

1 h = 60 min

1 min = 60 s

t= 1 h×60 min1 h×60 s1 min=3600 s

Plug the given values in the equation 1,

ln[X]t0.50 mol/L= -(6.32×10-4s-1×3600 s)

ln[X]t0.5 mol/L= - 2.2752

[X]t0.50 mol/L= e(- 2.2752)

[X]t0.50 mol/L= 0.1028

[X]t=0.1028× 0.50 mol/L

[X]t=0.051 mol/L

Therefore, the amount (concentration) of N2O5  remains in the solution after 1 h is,0.051 mol/L.

b) Percent of N2O5 reacted after 1 h:

In part (a) we have calculated the concentration of N2O5 remain unreacted after 1 h i.e. 0.051 mol/L, using this we can calculate the concentration of N2O5 reacted in 1 h.

Reacted concentration = Initial concentration  Unreacted concentration

Reacted concentration of N2O5=0.50 mol/L-0.051 mol/L

= 0.449 mol/L of N2O5

Use this value to calculate the percent of N2O5 reacted in 1 h.

Percent reacted= Reacted cocentrationInitial cocentration×100

Percent of N2O5 reacted= 0.449 mol/L0.50 mol/L×100

Percent of N2O5 reacted= 89.8 %

Percent of N2O5 reacted= 90 %(in correct significant figures)

Therefore, the percent of N2O5 reacted at 1 h is 90%

Conclusion:

We have used integrated rate law to find out the concentration of reactant at the given time after the reaction starts. From this we calculated the percent of reactant reacted in the reaction.

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Chapter 13 Solutions

Chemistry: An Atoms-Focused Approach

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