Chapter 16.1, Problem 3PP

Interpretation Introduction

Interpretation:

The concentration of $\text{HCl}$ after 4.00 s 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 3PP

Concentration of $\text{HCl}$ after 4.00 s is $\text{0}\text{.02}\text{mol}\text{/}\text{L}$.

Explanation of Solution

Given that $\text{HCl}$ is formed. So, the average rate expression should be positive.

$\begin{array}{c}\text{Average}\text{reaction}\text{rate=0}\text{.0050}\text{mol}\text{/}\text{L}\text{.s}\\ t_1=0.00s\\ t_2=4.00s\\ [\text{HCl}]\text{at}{t}_1=0.00M\end{array}$

$\begin{array}{c}\text{Average}\text{reaction}\text{rate=}\frac{\Delta [\text{HCl}]}{\Delta t}\\ 0.0050\text{mol}\text{/}(\text{L}·\text{s)}=\frac{[\text{HCl}]\text{at}\text{time}{t}_2-[\text{HCl}]\text{at}\text{time}{t}_1}{t_2-t_1}\\ [\text{HCl}]\text{at}\text{time}{t}_2=0.0050\text{mol}\text{/}(\text{L}·\text{s)}\times (t_2-t_1)+[\text{HCl}]\text{at}\text{time}{t}_1\\ =0.0050\text{mol}\text{/}(\text{L}·\text{s)}\times (4.00s-0.00s)+0.00\text{mol/}\text{L}\\ \text{=0}\text{.02}\text{mol}\text{/}\text{L}\end{array}$

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.