Chapter 14, Problem 103AP

The rate constant for the gaseous reaction:

${\text{H}}_{\text{2}}\text{(}g{\text{) + I}}_{\text{2}}\text{(}g\text{) }\to \text{ 2HI(}g\text{)}$

is $\text{2}\text{.42 }\times {\text{ 10}}^{\text{-2}}\text{/}M\cdot \text{ s at 400°C}\text{.}$ Initially an equimolar sample of ${\text{H}}_{\text{2}}$ and ${\text{I}}_{\text{2}}$ is placed in a vessel at $\text{400°C,}$ and the total pressure is 1658 mmHg. (a) What is the initial rate (M/min) of formation of HI? (b) Determine the rate of formation of HI and the concentration of HI (in molarity) after 10.0 min.

Interpretation Introduction

Interpretation:

The initial rate, the rate of formation and the concentration of $HI$ are to be determined.

Concept introduction:

The rate of reaction in terms of concentration of reactants is called the rate equation or rate law.

The ideal gas equation depicts a relation between four variables. As it predicts the state of gases, so it is called the equation of state or ideal gas equation.

The ideal gas equation is as follows:

$PV=n\text{R}T$

Here, $P$ is the pressure of the gas, $n$ is the number of moles, $V$ is the volume of gas, $\text{R}$ is gas constant, and $T$ is the absolute temperature.

Answer to Problem 103AP

Solution:

(a) $1.112\times {10}^{-3}M/\min$.

(b) $8.8\times {10}^{-3}M$.

Explanation of Solution

Given information: The given reaction is as follows:

${\text{H}}_{\text{2}}\left(g\right)+{\text{I}}_{\text{2}}\left(g\right)\to 2\text{HI}\left(g\right)$

The rate constant for this reaction is $2.42\times {10}^{-3}\text{}/\text{M}\text{.s}$. Initially, an equimolar sample of ${\text{H}}_{\text{2}}$ and ${\text{I}}_{\text{2}}$ is placed in a vessel at $400°\text{C}$ and at a total pressure of $5.42\text{ atm}$.

a) The initial rate (M/min) of formation of HI.

Based on the units of rate constant fora second-order reaction, the rate law for the reaction is as follows:

$\text{Rate}=\text{k}\left[{\text{H}}_2\right]\left[{\text{I}}_2\right]$

According to the ideal gas equation, the concentrations of $\left[H_2\right]$ and $\left[I_2\right]$ are determined below.

The rearranged ideal gas equation is as follows:

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

Substitute the values of $P$, $\text{R}$, and $T$ in the above equation.

In the givenreaction, there are equimolar amounts of hydrogen and iodine. Total pressure is $1658mmHg$ and the partial pressure for each gas is as follows:

$\begin{array}{c}{H}_2=I_2=\left(\frac{1658}{2}\right)\\ =829mmHg\end{array}$

One atmospheric is equal to $760mmHg$.

$\begin{array}{c}{H}_2=I_2=\left(\frac{829\cancel{mmHg}\left(\frac{1\cancel{atm}}{760\cancel{mmHg}}\right)}{0.0821L.\cancel{atm}/mol.\cancel{K}\left(673\cancel{K}\right)}\right)\\ =\frac{1.0907}{55.2533}\\ =0.01974M\end{array}$

The units of rate constant areconvertedinto $/M.min$ as follows:

$\begin{array}{c}k=\left(2.42\times {10}^{-2}\right)\left(\frac{1}{1M\sec }\right)\left(\frac{60s}{1\min }\right)\\ =1.452\frac{1}{M\min }\end{array}$

Substitute the rate constant in rate law as follows:

$\begin{array}{c}Rate=\left(1.452\frac{1}{\cancel{M}.\min }\right)\left[0.0197\cancel{M}\right]\left[0.01974M\right]\\ =\left(1.452{\min }^{-1}\times 0.000388M\right)\\ =0.0005635M/\min \\ =5.560\times {10}^{-4}M/\min \end{array}$

The rate is as follows:

$\begin{array}{c}\text{Rate}=\frac{1}{2}\frac{\Delta \left[\text{HI}\right]}{\Delta t}\\ \frac{\Delta \left[\text{HI}\right]}{\Delta t}=\left(2\times \text{rate}\right)\\ =2\left(5.635\times {10}^{-4}\text{M}/\min \right)\\ =1.13\times {10}^{-3}\text{M}/\min \end{array}$

The rate is $1.13\times {10}^{-3}M/\min$.

b) The rate of formation of HI and the concentration of HI (in morality) after 10.0 min.

According to the second order rate law, the concentration of $H_2$ after $10\min$ is as follows:

$\frac{1}{{\left[\text{A}\right]}_t}=\text{k}t+\frac{1}{{\left[\text{A}\right]}_0}$

Here, ${\left[\text{A}\right]}_t$ is the final concentration of reactant, ${\left[\text{A}\right]}_0$ is the initial concentration of reactant, $\text{k}$ is rate constant, and $t$ is time.

Substitute the values ${\left[\text{A}\right]}_0$, $\text{k}$, and $t$ in the aboveequation as follows:

$\begin{array}{c}\frac{1}{{\left[H\right]}_t}=\left(1.452\frac{1}{M.\cancel{\min }}\right)\left(10.0\cancel{\min }\right)+\left(\frac{1}{0.01974M}\right)\\ =65.178\\ {\left[H\right]}_t=\left(\frac{1}{65.178}\right)\\ =0.01534M\end{array}$

After $10\min$, the concentration of iodine also equals the concentration of hydrogen $\left(H_2\right)$ as follows:

$\text{Rate}=\text{k}\left[{\text{H}}_2\right]\left[{\text{I}}_2\right]$

Substitute the values of concentration of hydrogen, iodine and rate constant in the above expression,

$\begin{array}{c}Rate=\left(1.452\frac{1}{\min }\right)\left[0.01534M\right]\left[0.01534M\right]\\ =\left(1.452\right)\left(0.0002353\right)\\ =0.0003417\\ =3.417\times {10}^{-4}M/\min \end{array}$

The rate is calculated as follows:

$\begin{array}{l}\text{Rate}=\frac{1}{2}\frac{\Delta \left[\text{HI}\right]}{\Delta t}\\ \frac{\Delta \left[\text{HI}\right]}{\Delta t}=2\times \text{rate}\\ =2\left(3.416\times {10}^{-4}\text{M}/\min \right)\\ =6.832\times {10}^{-4}\text{M}/\min \end{array}$

The concentration of $\left[HI\right]$ after $10\min$ is as follows:

$\begin{array}{c}{\left[HI\right]}_t=\left({\left[H_2\right]}_0-{\left[H_2\right]}_t\right)\times 2\\ =\left(0.01974M-0.01534M\right)2\\ =\left(0.0044\right)2\\ =0.0088\text{ or }8.8\times {10}^{-3}M\end{array}$

The concentration of $\left[HI\right]$ after $10\min$ is $8.8\times {10}^{-3}M$.