Chapter 13, Problem 13.132QP

A gas mixture containing CH₃ fragments, C₂H₆ molecules, and an inert gas (He) was prepared at 600 K with a total pressure of 5.42 atm. The elementary reaction

${\text{CH}}_3+{\text{C}}_2{\text{H}}_6\stackrel{}{\to }{\text{CH}}_4+{\text{C}}_2{\text{H}}_5$

has a second-order rate constant of 3.0 × 10⁴/M · s. Given that the mole fractions of CH₃ and C₂H₆ are 0.00093 and 0.00077, respectively, calculate the initial rate of the reaction at this temperature.

Interpretation Introduction

Interpretation:

The initial rate of the given reaction at the given temperature has to be calculated.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Rate constants are independent of concentration but depend on other factors, most notably temperature.

The reaction with the faster rate will have the larger rate constant.

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

The partial pressure of a gas in a mixture can be expressed as,

$\begin{array}{l}{\text{P}}_{\text{A}}={\text{X}}_{\text{A}}{\text{P}}_{\text{Total}}\\ {\text{P}}_{\text{A}}\to \text{Partial}\text{pressure}\text{of}\text{gas}\text{A}\\ {\text{X}}_{\text{A}}\to \text{Mole}\text{fraction}\text{of}\text{gas}\text{A}\\ {\text{P}}_{\text{Total}}\to \text{Total}\text{pressure}\text{of}\text{mixture}\end{array}$

The ideal gas Law equation is,

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{P}\to \text{Pressure}\\ \text{V}\to \text{Volume}\\ \text{n}\to \text{Number}\text{of}\text{moles}\\ \text{R}\to \text{Ideal}\text{gas}\text{constant}\\ \text{T}\to \text{Temperature}\end{array}$

Answer to Problem 13.132QP

Initial rate of the given reaction is $2.6\times 10^{-4}M/s$

Explanation of Solution

Given,

Gas mixture containing $C{H}_3$ fragments, $C_2{H}_6$ molecules, and an inert gas $\left(He\right)$ was prepared at $600K$ with a total pressure of $5.42atm$.

The elementary reaction is,

$C{H}_3+C_2{H}_6\to C{H}_4+C_2{H}_5$

This reaction follows second order kinetics, with a rate constant of $\text{3}\text{.0}\times {\text{10}}^{\text{4}}\text{/}\text{M}\text{.s}$ .

Given mole fractions of $C{H}_3$ and $C_2{H}_6$ are $0.00093and0.00077$ respectively,

The initial rate of the reaction at $600K$ temperature can be determined as follows,

Rate law for the given reaction is,

$Rate=k\left[C{H}_3\right]\left[C_2{H}_6\right]$

Rate constant value for the given reaction is $\text{3}\text{.0}\times {\text{10}}^{\text{4}}\text{/}\text{M}\text{.s}$. If the concentration of reactants were known it is easy to find out the initial rate of the reaction.

Using mole fraction value and total partial pressure, partial pressure of each reactant in the reaction can be calculated as follows,

$\begin{array}{l}{P}_{C{H}_3}=X_{C{H}_3}{P}_T=\left(0.00093\right)\left(5.42atm\right)=0.0050atm\\ P_{C_2{H}_6}=X_{C_2{H}_6}{P}_T=\left(0.00077\right)\left(5.42atm\right)=0.0042atm\end{array}$

Molar concentration of reactants can be determined with the help of ideal gas equation,

$\begin{array}{l}\frac{n}{V}=\frac{P}{RT}\\ \frac{n}{V}\to molarconcentration\end{array}$

$\begin{array}{l}{M}_{C{H}_3}=\frac{P_{C{H}_3}}{RT}=\frac{0.00050atm}{0.0821L.atm/mol.K\times 600K}=1.0\times {10}^{-4}M\\ M_{C_2{H}_6}=\frac{P_{C_2{H}_6}}{RT}=\frac{0.00042atm}{0.0821L.atm/mol.K\times 600K}=8.5\times {10}^{-5}M\\ \end{array}$

Substitute the concentration and the rate constant into the rate law to find the initial rate of the reaction,

$\begin{array}{l}\text{Rate}=\text{k}\left[{\text{CH}}_{\text{3}}\right]\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]\\ \text{Rate}=\left(\text{3}\text{.0}{\text{×10}}^{\text{4}}{\text{M}}^{\text{-1}}{\text{s}}^{\text{-1}}\right)\text{×}\left(\text{1}\text{.0}\text{×}{\text{10}}^{\text{-4}}\text{M}\right)\text{×}\left(\text{8}\text{.5}\text{×}{\text{10}}^{\text{-5}}\text{M}\right)\\ Rate=2.6\times 10^{-4}M/s\end{array}$