Chapter 16.1, Problem 1PP

Interpretation Introduction

Interpretation: The average reaction rate expressed in moles ${\text{H}}_{\text{2}}$ consumed per litre per second needs to be determined.

Concept introduction: The average rate of an action or process is defined as the change in a given quantity during a specific period of time. In equation form, the average rate is expressed as follows:

  $\text{Average}\text{rate=}\frac{\Delta \text{Quantity}}{\Delta t}$

Answer to Problem 1PP

The average reaction rate if ${\text{H}}_{\text{2}}$ is consumed is $0.0025\text{mol}/(\text{L}·\text{s})$ .

Explanation of Solution

Given that ${\text{H}}_{\text{2}}$ is consumed. This indicates the average rate of expression should be negative.

Given values are,

  $\begin{array}{c}{\text{[H}}_{\text{2}}]at{t}_1=0.030\text{M}\\ {\text{[H}}_2\text{]}\text{at}{t}_2=0.020\text{M}\\ t_1=0.00s\\ t_2=4.00s\end{array}$

So, the expression of the average reaction rate is,

  $\begin{array}{c}\text{Average}\text{reaction}\text{rate=}-\frac{\Delta [{\text{H}}_{\text{2}}]}{\Delta t}\\ =-\frac{[{\text{H}}_{\text{2}}]\text{at}\text{time}{t}_2-[{\text{H}}_{\text{2}}]\text{at}\text{time}{t}_1}{t_2-t_1}\\ =-\frac{0.020\text{M}-0.030\text{M}}{4.00s-0.00s}\\ =\frac{0.010\text{M}}{4.00s}\\ =0.0025\text{mol}/(\text{L}·\text{s})\end{array}$

Conclusion

The negative sign in the reaction rate expression indicates that the concentration of the compound decreases as the reaction proceeds but the reaction rate is always positive.