Chapter 13, Problem 13.69QA

Interpretation Introduction

To find:

a) In the first order reaction of decomposition of $N_2{O}_5$ whatamount (concentration) of $N_2{O}_5$ remains in the solution after $1 h$ if the initial concentration was $0.50 mol/$ L and $k=6.32\times {10}^{-4}{s}^{-1}$?

b) Percentage of $N_2{O}_5$ reacted after $1 h$.

Answer to Problem 13.69QA

Solution:

a) The amount (concentration) of $N_2{O}_5$ remains in the solution after $1 h$ is,$0.051 mol/L$.

b) Percent of $N_2{O}_5$ 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

$\ln \frac{{\left[X\right]}_t}{{\left[X\right]}_0}= -kt$ …(equation 1)

i.e. $ln {\left[X\right]}_t= -kt+{ln \left[X\right]}_0$

Where

${\left[X\right]}_t$= concentration of reactant at time t

${\left[X\right]}_0$= initial concentration of reactant

$k$= rate constant

$t$= reaction time

ii. Percent reacted:

$Reacted concentration = Initial concentration – Unreacted concentration$

$Percent reacted= \frac{Reacted cocentration}{Initial cocentration}\times 100$

3) Given:

${\left[X\right]}_0=0.50 mol/L$

$k=6.32\times {10}^{-4}{s}^{-1}$

$t=1 h$

${\left[X\right]}_t= ?$

4) Calculation:

a) Amount (concentration) of $N_2{O}_5$ remains in the solution after $1 h$:

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

$1 h = 60 min$

$1 min = 60 s$

$t= 1 h\times \frac{60 min}{1 h}\times \frac{60 s}{1 min}=3600 s$

Plug the given values in the equation 1,

$\ln \frac{{\left[X\right]}_t}{0.50 mol/L}= -(6.32\times {10}^{-4}{s}^{-1}\times 3600 s)$

$\ln \frac{{\left[X\right]}_t}{0.5 mol/L}= - 2.2752$

$\frac{{\left[X\right]}_t}{0.50 mol/L}= e^{(- 2.2752)}$

$\frac{{\left[X\right]}_t}{0.50 mol/L}= 0.1028$

${\left[X\right]}_t=0.1028\times 0.50 mol/L$

${\left[X\right]}_t=0.051 mol/L$

Therefore, the amount (concentration) of $N_2{O}_5$ remains in the solution after $1 h$ is,$0.051 mol/L$.

b) Percent of $N_2{O}_5$ reacted after $1 h$:

In part (a) we have calculated the concentration of $N_2{O}_5$ remain unreacted after $1 h$ i.e.$0.051 mol/L$, using this we can calculate the concentration of $N_2{O}_5$ reacted in $1 h$.

$Reacted concentration = Initial concentration – Unreacted concentration$

$Reacted concentration of N_2{O}_5=0.50 mol/L-0.051 mol/L$

$= 0.449 mol/L of N_2{O}_5$

Use this value to calculate the percent of $N_2{O}_5$ reacted in $1 h$.

$Percent reacted= \frac{Reacted cocentration}{Initial cocentration}\times 100$

$Percent of N_2{O}_5 reacted= \frac{0.449 mol/L}{0.50 mol/L}\times 100$

$Percent of N_2{O}_5 reacted= 89.8 \%$

$Percent of N_2{O}_5 reacted= 90 \%$(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.