Chapter 16, Problem 82A

Interpretation Introduction

Interpretation:

To determine the units of rate and rate constant of third-order reaction.

Concept introduction:

Consider a general reaction

$\text{aA+bB}\to \text{Products}$

The rate law of the above reaction is

$\text{rate}=k{\left[\text{A}\right]}^m{\left[\text{B}\right]}^n$

Here, $\left[\text{A}\right]$ and $\left[\text{B}\right]$ is the concentration of reactant A and B respectively, k represents specific rate and m and n represents order of corresponding reactants. Overall order of the reaction is $m+n$.

Answer to Problem 82A

For third-order reaction units of rate is $\text{mol/L}\text{.s}$ and rate constant is $\frac{{\text{L}}^{\text{2}}\text{.s}}{{\text{mol}}^{\text{2}}}$.

Explanation of Solution

Consider a general reaction

$\text{aA+bB}\to \text{Products}$

The rate law of the above reaction is

$\text{rate}=k{\left[\text{A}\right]}^m{\left[\text{B}\right]}^n$

For a third order reaction $m+n=3$. Concentration of each reactant has units $\text{mol/L}$. Substitute the units into the rate law.

$\begin{array}{c}\text{rate}=k{\left[\text{A}\right]}^m{\left[\text{B}\right]}^n\\ =k{\left[\text{mol/L}\right]}^2{\left[\text{mol/L}\right]}^1\\ =k\left[{\text{mol}}^3{\text{/L}}^3\right]\end{array}$

The unit of rate of a reaction is $\text{mol/L}\text{.s}$. Plug the units of rate in above expression.

$\begin{array}{c}\text{rate}=k\left[{\text{mol}}^{\text{3}}{\text{/L}}^{\text{3}}\right]\\ \text{mol/L}\text{.s}=k\left[{\text{mol}}^{\text{3}}{\text{/L}}^{\text{3}}\right]\\ k=\frac{\text{mol/L}\text{.s}}{{\text{mol}}^{\text{3}}{\text{/L}}^{\text{3}}}\\ =\frac{{\text{L}}^{\text{2}}\text{.s}}{{\text{mol}}^{\text{2}}}\end{array}$

Therefore, units of rate is $\text{mol/L}\text{.s}$ and rate constant is $\frac{{\text{L}}^{\text{2}}\text{.s}}{{\text{mol}}^{\text{2}}}$.

Conclusion

For third-order reaction units of rate is $\text{mol/L}\text{.s}$ and rate constant is $\frac{{\text{L}}^{\text{2}}\text{.s}}{{\text{mol}}^{\text{2}}}$.