Chapter 19, Problem 19.131QP

The rale constant for the gaseous reaction

H₂(g) + I₂(g) → 2HI(g)

is 2.42 × 10⁻²/M · s at 400°C. Initially an equimolar sample of H₂ and I₂ is placed in a vessel at 400°C, and the total pressure is 1658 mmHg. (a) What is the initial rate (M/min) of formation of HI? (b) What are the rate of formation of HI and the concentration of HI (in molarity) after 10.0 min?

Interpretation Introduction

Interpretation:

The initial rate of formation of $\text{HI}$ has to be determined.  The rate formation of $\text{HI}$ and the concentration of $\text{HI}$ in molarity after 10 min has to be determined.

Concept introduction:

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

• The change in concentration term is divided by the respective stoichiometric coefficient.
• The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
• Rate of reaction is always represented by positive quantities.

Answer to Problem 19.131QP

The initial rate of formation of $\text{HI}$ is $1.13\times {10}^{-3}M/\min$

The rate formation of $\text{HI}$ and the concentration of $\text{HI}$ in molarity after 10 min is $8.80\times {10}^{-3}M$

Explanation of Solution

(a)

From the given information, the units of rate constant denotes a second-order reaction and the rate law is probably as follows

${\text{rate = k[H}}_{\text{2}}{\text{][I}}_{\text{2}}\text{]}$

By using the ideal gas equation to solve for the initial concentrations of ${\text{H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}$.  Then to calculate initial rate with respect to ${\text{H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}$ and then convert it into the initial rate of formation of $\text{HI}$ and carryout the significant figure throughout this calculation to minimize rounding errors

$n=\frac{PV}{RT}$

$\frac{n}{V}=M=\frac{P}{RT}$

The total pressure is 1658 mmHg and there are equimolar amounts of ${\text{H}}_{\text{2}}\text{and}{\text{I}}_{\text{2}}$ in the vessel and the partial pressure of each gas could be 829 mmHg.

$\left[{\text{H}}_{\text{2}}\right]\text{=}\left[{\text{I}}_{\text{2}}\right]\text{=}\frac{\left(\text{829}\text{mmHg×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.atm}}{\text{K}\text{.atm}}\right)\left(\text{400+273}\right)\text{K}}=0.01974M$

Now, let’s convert the units of the rate constant to /M.min, and substitute the corresponding values in the rate law to solve the rate as follows.

$\text{k = 2}{\text{.42×10}}^{\text{-2}}\frac{\text{1}}{\text{M}\text{.s}}\text{×}\frac{\text{60s}}{\text{1min}}\text{=1}\text{.452}\frac{\text{1}}{\text{M}\text{.min}}$

${\text{rate = k[H}}_{\text{2}}{\text{][I}}_{\text{2}}\text{]}$

$\text{rate =}\left(\text{ 1}\text{.452}\frac{\text{1}}{\text{M}\text{.min}}\right)\left(0.01974M\right)\left(0.01974\right)=5.658\times {10}^{-4}M/\min$

We know that,

$\text{rate}\text{=}\frac{\text{1}}{\text{2}}\frac{\text{Δ}\left[\text{HI}\right]}{\text{Δt}}$

or

$\frac{\text{Δ}\left[\text{HI}\right]}{\text{Δt}}=2\text{×rate = }\left(\text{2}\right)\left(5.658\times {10}^{-4}M/\min \right)=1.13\times {10}^{-3}M/\min$

(b)

The integrated second-order rate law is used to calculate the concentration of ${\text{H}}_{\text{2}}$ after 10.0min.  Then substitute the obtained concentration into the rate law to get rate.

$\begin{array}{l}\frac{\text{1}}{{\left[{\text{H}}_{\text{2}}\right]}_{\text{t}}}\text{=}\left(\text{1}\text{.452}\frac{\text{1}}{\text{M}\text{.min}}\right)\left(\text{10}\text{.0}\text{min}\right)\text{+}\frac{\text{1}}{\text{0}\text{.01974}\text{M}}\\ \\ {\left[{\text{H}}_{\text{2}}\right]}_{\text{t}}=0.01534M\end{array}$

Similarly we can calculate the concentration of ${\text{I}}_{\text{2}}$ after 10.0min and it will also equal to

${\left[{\text{I}}_{\text{2}}\right]}_{\text{t}}\text{= 0}\text{.01534}M$

Now,

${\text{rate = k[H}}_{\text{2}}{\text{][I}}_{\text{2}}\text{]}$

$\text{rate = }\left(\text{ 1}\text{.452}\frac{\text{1}}{\text{M}\text{.min}}\right)\left(0.01534M\right)\left(0.01534M\right)=3.417\times {10}^{-4}M/\min$

We know that,

$\frac{\text{Δ}\left[\text{HI}\right]}{\text{Δt}}\text{= 2 × rate }\text{=}\left(\text{2}\right)\left(3.417\times {10}^{-4}M/\min \right)=6.83\times {10}^{-4}M/\min$

Therefore, the concentration of $\text{HI}$ after 10.0 min is

${\text{[HI]}}_{\text{t}}{\text{ = ([H}}_{\text{2}}{\text{]}}_{\text{0}}{\text{- [H}}_{\text{2}}{\text{]}}_{\text{t}}\text{) }\times \text{2}$

${\text{[HI]}}_{\text{t}}\text{ =}\left(\text{0}\text{.01974 M- 0}\text{.01534 M}\right)\times 2=8.80\times {10}^{-3}M$

Conclusion

The initial rate of formation of $\text{HI}$ was $1.13\times {10}^{-3}M/\min$.  The rate formation of $\text{HI}$ and the concentration of $\text{HI}$ in molarity after 10 min was $8.80\times {10}^{-3}M$.