Chapter 14, Problem 104AP

A gas mixture containing ${\text{CH}}_{\text{3}}$ fragments. ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ molecules, and an inert gas (He) was prepared at 600 K with a total pressure of 5.42 atm. The elementary reaction

${\text{CH}}_{\text{3}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{6}}\to {\text{CH}}_{\text{4}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{5}}$

has a second-order rate constant of $\text{3}{\text{.0 x 10}}^{\text{4}}\text{/M}\cdot \text{S}\text{.}$ Given that the mole fractions of ${\text{CH}}_{\text{3}}$ at ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ 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 reaction at given temperature is to be calculated.

Concept introduction:

The branch of chemistry thatdeals with the relative rates of chemical reactions and the factors affecting the rates of reactions is known as chemical kinetics.

The mathematical expression thatdefines the dependence of the rate of reaction on the concentration of reactants is known as the rate law equation or the rate law.

Answer to Problem 104AP

Solution: $2.6\times {10}^{-4}M/s$.

Explanation of Solution

Given information: The reaction is given as follows:

${\text{CH}}_3+{\text{C}}_{\text{2}}{\text{H}}_6\to {\text{CH}}_4+{\text{C}}_{\text{2}}{\text{H}}_5$

The rate constant of the reaction is $3.0\times {10}^4{M}^{-1}{s}^{-1}$.

The total pressure of the system is $5.42atm$.

The partial pressure of methyl molecule is calculated by the expression as follows:

$P_{C{H}_3}=X_{C{H}_3}{P}_T$

Here, $P_{C{H}_3}$ is the partial pressure of the methyl group, $X_{C{H}_3}$ is the mole fraction of the methyl group, and $P_T$ is the total pressure.

Substitute the values of $X_{C{H}_3}$ and $P_T$ in the above equation,

$\begin{array}{c}{P}_{C{H}_3}=\left(0.00093\right)\left(5.42\text{atm}\right)\\ =0.0050\text{atm}\end{array}$

The partial pressure of ethane molecule is calculated by the expression as follows:

$P_{C_2{H}_6}=X_{C_2{H}_6}{P}_T$

Here, $P_{C_2{H}_6}$ is the partial pressure of ethyl group, $X_{C_2{H}_6}$ is the mole fraction of ethyl group, and $P_T$ is the total pressure.

Substitute the values of $X_{C_2{H}_6}$ and $P_T$ in the above equation,

$\begin{array}{c}{P}_{C_2{H}_6}=\left(0.00077\right)\left(5.42\text{atm}\right)\\ =0.0042\text{atm}\end{array}$ a

According to the ideal gas equation:

$\begin{array}{c}PV=n\text{R}T\\ \frac{n}{V}=\frac{P}{\text{R}T}\left(\therefore M=\frac{n}{V}\right)\\ M=\frac{P}{\text{R}T}\end{array}$

Here, $n$ is the number of moles, $V$ is the volume, $P$ is the pressure, $R$ is the universal gas constant, $M$ is the molarity, and $T$ is the absolute temperature.

The molarity concentration of ${\text{CH}}_3$ is calculated by the expression as follows:

$M_{C{H}_3}=\frac{P_{C{H}_3}}{\text{R}T}$

Here, $M_{C{H}_3}$ is the molarity of methyl, $P_{C{H}_3}$ is the partial pressure of methyl group, $\text{R}$ is the gas constant, and $T$ is the absolute temperature.

Substitute the values of $P_{C{H}_3}$, $\text{R}$, and $T$ in the above equation,

$\begin{array}{c}{M}_{C{H}_3}=\frac{0.0050\cancel{\text{atm}}}{\left(0.0821\text{L}\text{.}\cancel{\text{atm}}/\text{mol}\text{.}\cancel{\text{K}}\right)\left(600\bcancel{\text{K}}\right)}\\ =\frac{0.0050}{49.26}\text{M}\\ =1.01\times {10}^{-4}\text{M}\end{array}$

The molarity concentration of ${\text{C}}_2{\text{H}}_6$ is calculated by the expression as follows:

$M_{{\text{C}}_2{\text{H}}_6}=\frac{P_{{\text{C}}_2{\text{H}}_6}}{RT}$

Here, $M_{{\text{C}}_2{\text{H}}_6}$ is the molarity concentration of ethane, $P_{{\text{C}}_2{\text{H}}_6}$ is the partial pressure of ethane, $R$ is the gas constant, and $T$ is the absolute temperature.

Substitute the values of $P_{{\text{C}}_2{\text{H}}_6}$, $R$, and $T$ in the above equation,

$\begin{array}{c}{M}_{{\text{C}}_2{\text{H}}_6}=\frac{0.0042\cancel{\text{atm}}}{\left(0.0821\text{L}\text{.}\cancel{\text{atm}}/\text{mol}\text{.}\cancel{\text{K}}\right)\left(600\bcancel{\text{K}}\right)}\\ =\frac{0.0042}{49.26}\text{M}\\ =8.5\times {10}^{-5}\text{M}\end{array}$

The rate of the reaction is calculated by the expression as follows:

$\text{rate}=k\left[{\text{CH}}_3\right]\left[{\text{C}}_2{\text{H}}_6\right]$

Substitute the values of $k,\left[{\text{CH}}_3\right]$, and $\left[{\text{C}}_2{\text{H}}_6\right]$ in the above equation,

$\begin{array}{c}\text{rate}=\left(3.0\times {10}^4\cancel{{\text{M}}^{-1}}{\text{s}}^{-1}\right)\left(1.01\times {10}^{-4}\cancel{\text{M}}\right)\left(8.5\times {10}^{-5}\text{M}\right)\\ =25.5\times {10}^{-5}\text{M}/\text{s}\\ =2.6\times {10}^{-4}\text{M}/\text{s}\end{array}$

Conclusion

The initial rate of the reaction is $2.6\times {10}^{-4}M/s$.