Chapter 16, Problem 12STP

Interpretation Introduction

Interpretation:

The instantaneous reaction rate needs to be explained.

Concept introduction:

A rate law represents the relation between a chemical reaction rate and reactant concentration. The rate law for the one step reaction $\text{A}\to \text{B}$ is as follows:

$\text{rate=}k\left[\text{A}\right]$

Here, k and $\left[\text{A}\right]$ represents rate constant and concentration of reactant respectively.

Answer to Problem 12STP

Instantaneous reaction rate ( k ) is $8.774\times {10}^{-7}\times \frac{\text{mol}}{\text{L}\text{.s}}$.

Explanation of Solution

The rate law for the reaction $\text{A+B+C}\to \text{Products}$ is given below.

$\text{rate}=k\left[\text{A}\right]\left[\text{B}\right]\left[\text{C}\right]$ …… (1)

Given values are as follows:

$\begin{array}{l}k=6.92\times {10}^{-5}{\text{L}}^{\text{2}}\text{/}\left({\text{mol}}^{\text{2}}\text{.s}\right)\\ \left[\text{A}\right]=0.175\text{M}\\ \left[\text{B}\right]\text{=0}\text{.230M}\\ \left[\text{C}\right]=0.315\text{M}\end{array}$

Therefore, instantaneous reaction rate ( k ) is calculated using the following formula:

Put all values in equation (1).

$\begin{array}{c}\text{Rate}=k\left[\text{A}\right]\left[\text{B}\right]\left[\text{C}\right]\\ =6.92\times {10}^{-5}{\text{L}}^{\text{2}}\text{/}\left({\text{mol}}^{\text{2}}\text{.s}\right)\times 0.175\frac{\text{mol}}{\text{L}}\text{0}\text{.230}\times \frac{\text{mol}}{\text{L}}\times 0.315\times \frac{\text{mol}}{\text{L}}\\ =0.08774\times {10}^{-5}\times \frac{\text{mol}}{\text{L}\text{.s}}\\ =8.774\times {10}^{-7}\times \frac{\text{mol}}{\text{L}\text{.s}}\end{array}$

Conclusion

Instantaneous reaction rate ( k ) is $8.774\times {10}^{-7}\times \frac{\text{mol}}{\text{L}\text{.s}}$.