Chapter 13, Problem 13.70QA

Interpretation Introduction

To find:

What percent of ${PH}_3$ reacts in $1 \mathrm{m}\mathrm{i}\mathrm{n}$ in the decomposition reaction at ${600}^{o}C$ with $k=0.023 s^{-1}$ and initial partial pressure of ${PH}_3$ is $375 torr$, where the reaction is first order?

Answer to Problem 13.70QA

Solution:

The percent of ${PH}_3$ reacts in $1 min$ in the decomposition reaction is $75\%$.

Explanation of Solution

1) Concept:

In first order reaction, the rate nearly depends on only one reactant, so it is possible to find out the concentration of reactant at any time once the reaction get started and hence percent reacted of reactant can also be find out using integrated rate law.

As the unit concentration in terms $\ln {({\left[X\right]}_t/[X]}_0)$ cancel out in integrated rate law for the first order reactions, hence the molar concentration can be replaced by any concentration term.

2) Formula:

i. Integrated rate law for first order reaction is,

$\ln \frac{{\left[X\right]}_t}{{\left[X\right]}_0}= -kt$

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

In the given question, concentration term is partial pressure.

Consider the initial partial pressure of ${PH}_3$ is $P_1$ and at time $t$ is, $P_2$, the integrated rate law becomes,

$\ln \frac{P_2}{P_1}= -kt$ …(equation 1)

ii. Percent reacted:

$pressure used = Initial partial pressure – Unreacted partial pressure$

$Percent reacted= \frac{Reacted partial pressure}{Initial Partial pressure}\times 100$

3) Given:

$P_1=375 torr$

$k=0.023 s^{-1}$

$t=1 min$

4) Calculation:

a. Amount of ${PH}_3$ remains in the solution after $1 m$ in terms of pressure :

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

$t=1 min = 60 s$

Plug the given values in the equation 1,

$\ln \frac{P_2}{375 torr}= -(0.023 s^{-1}\times 60 s)$

$\ln \frac{P_2}{375 torr}= - 1.38$

$\ln \frac{P_2}{375 torr}= e^{(- 1.38)}$

$\frac{P_2}{375 torr}= 0.2516$

$P_2=0.2516\times 375 torr$

$P_2=94 torr$

Therefore, the partial pressure of ${PH}_3$ after $1 min$ is,$94 torr$.

b. Percent of ${PH}_3$ reacted after $1 min$:

In part (a), we got the partial pressure of unreacted ${PH}_3$ after $1 min$ i.e.$94 torr$, using this we can calculate the partial pressure of reacted ${PH}_3$ in $1 min$.

$Partial pressure of reacted {PH}_3= Initial partial pressure – partial pressure of unreacted {PH}_3$

$Partial pressure of reacted {PH}_3=375 torr-94 torr$

$Partial pressure of reacted {PH}_3=281 torr$

Use this value to calculate the percent of ${PH}_3$ reacted in $1 min$.

$Percent of {PH}_3 reacted= \frac{Partial pressure of reacted {PH}_3}{Initial partial pressure}\times 100$

$Percent of {PH}_{3}reacted= \frac{281 torr}{375 torr}\times 100$

$Percent of {PH}_3 reacted= 74.9 \%$

$Percent of {PH}_{3}reacted= 75 \%$(in correct significant figures)

Conclusion:

We used integrated rate law to find out the concentration of reactant at given time after the reaction starts and calculated the percent of PH₃ reacted in reaction.