Chapter 16, Problem 17STP

Interpretation Introduction

Interpretation:

The correct answer regarding the concentrations of ${\text{SO}}_{\text{2}}$ and ${\text{Cl}}_{\text{2}}$ at 200.0 min 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 17STP

Option B 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 concentrations of ${\text{SO}}_{\text{2}}$ and ${\text{Cl}}_{\text{2}}$ at 200.0 min as follows:

$\begin{array}{c}\text{Concentration}\text{of}\text{product=average}\text{reaction rate}\times \text{number of minutes passed}\\ \text{=1}\text{.3}\times {\text{10}}^{-3}\frac{\text{mol}}{\text{L}\text{.min}}\times 200\min \\ =2.6\times {10}^{-1}\text{mol/L}\\ \text{=0}\text{.26}\text{mol/L}\end{array}$

Therefore, concentration of the each product is $\text{0}\text{.26}$ M.

Conclusion

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