Chapter 16, Problem 65A

Interpretation Introduction

Interpretation:

To explain the change in the reaction rate law of rate ${\left[\text{A}\right]}^{\text{2}}{\left[\text{B}\right]}^{\text{3}}$ and the reaction rate is $4.5\times {10}^{-4}\text{mol/}\left(\text{L}\text{.min}\right)$.

Concept introduction:

Consider the general reaction

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

Rate law of the reaction is as follows:

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

Here, $\left[\text{A}\right]$, $\left[\text{B}\right]$ concentration of reactant A and B, k represents rate and m, n are the orders of their corresponding reactant.

Answer to Problem 65A

New reaction rate is $1.44\times {10}^{-2}\text{mol/L}\text{.min}$.

Explanation of Solution

Given rate law of the reaction is as follows:

$\text{rate=}{\left[\text{A}\right]}^2{\left[\text{B}\right]}^3$

The exponents 2 and 3 are the order with respect to reactant A and B. When the concentration of A doubles the rate increased by $2^2=4$ and when the concentration of B doubles, rate is increase by $2^3=8$. If the concentration of both A and B double, then rate is $4\times 8=32$ that means rate is increased by 32 times.

Calculate new rate as follows:

Initial rate is $4.5\times {10}^{-4}\text{mol/L}\text{.min}$.

$\begin{array}{c}\text{New}\text{rate=32}\times 4.5\times {10}^{-4}\text{mol/L}\text{.min}\\ \text{=144}\times {10}^{-4}\text{mol/L}\text{.min}\\ =1.44\times {10}^{-2}\text{mol/L}\text{.min}\end{array}$

Therefore, new rate is $1.44\times {10}^{-2}\text{mol/L}\text{.min}$.

Conclusion

From the above solution it is clear that rates of a reaction increased with increase in the reactant concentration.