Chapter 16, Problem 18STP

Interpretation Introduction

Interpretation:

The time taken for the decomposition of half of the original amount of the ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at average reaction rate needs to be determined.

Concept introduction:

Average reaction rate is calculated by using following formula

$\text{Average}\text{reaction}\text{rate=}-\frac{\Delta \left[\text{reactant}\right]}{\Delta t}$

Here, $\Delta \left[\text{reactant}\right]$ and $\Delta t$ represent the reactant concentration and change in time.

The negative sign in front of the above formula indicates the decrease in concentration of the reactant to yield a positive average reaction rate.

Answer to Problem 18STP

Option C is the correct answer.

Explanation of Solution

Experimental data of the reaction ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{SO}}_{\text{2}}\left(\text{g}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)$ is as follows:

Time (min)	$\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]\left(\text{M}\right)$	$\left[{\text{SO}}_{\text{2}}\right]\left(\text{M}\right)$	$\left[{\text{Cl}}_{\text{2}}\right]\left(\text{M}\right)$
$0.0$	$1.00$	$0.0$	$0.0$
$100.0$	$0.87$	$0.13$	$0.13$
$200.0$	$0.74$	?	?

$\text{Average}\text{reaction}\text{rate=}-\frac{\Delta \left[\text{reactant}\right]}{\Delta t}$ …… (1)

Here, $\Delta \left[\text{reactant}\right]$ and $\Delta t$ represent the reactant concentration and change in time.

Here, reactant is ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ use the table for first 100 minutes to calculate the reaction rate. The change in reactant concentration is

$\Delta \left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]=0.87\text{M}-1.00\text{M}$

And change in time is-

$\Delta t=100.0\min -0.0\min$

Put these values in equation (1).

$\begin{array}{c}\text{Average}\text{reaction}\text{rate=}-\frac{\Delta \left[\text{reactant}\right]}{\Delta t}\\ =-\frac{\Delta \left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}{\Delta t}\\ =-\frac{0.87\text{M}-1.00\text{M}}{100.0\min -0.0\min }\\ =-\frac{-0.13\frac{\text{mol}}{\text{L}}}{100\min }\end{array}$

$\begin{array}{c}\text{Average}\text{reaction}\text{rate}=-\frac{-0.13\frac{\text{mol}}{\text{L}}}{100\min }\\ =0.13\times {10}^{-2}\text{mol/L}\text{.min}\\ \text{=1}\text{.3}\times {10}^{-3}\text{mol/L}\text{.min}\end{array}$

Therefore, the average reaction rate is $\text{1}\text{.3}\times {10}^{-3}\text{mol/L}\text{.min}$.

Calculate the time taken for decomposition of half of the original amount of the ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at average reaction rate is as follows:

Half of the initial reactant rate $\frac{1.00}{2}=0.50\text{M}$

Therefore,

$\begin{array}{c}\frac{0.50\text{M}}{1.3\times {10}^{-3}\frac{\text{mol}}{\text{L}\text{.min}}}=\frac{0.50\frac{\text{mol}}{\text{L}}}{1.3\times {10}^{-3}\frac{\text{mol}}{\text{L}\text{.min}}}\\ =0.385\times {10}^{-3}\min \\ =385\min \end{array}$

Hence, time take is $385\min$.

Conclusion

The calculation shows that the correct option is C. Therefore, option A, B, D and E are not correct.