Chapter 16, Problem 63A

Interpretation Introduction

Interpretation:

To explain the change in the reaction rate of $\text{A+B}\to \text{AB}$ when the concentration of A is halved and all other conditions remain unchanged.

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 63A

If the concentration of reactant A is halved and all other circumstances remain constant in the reaction $\text{A+B}\to \text{AB}$ then the new rate will be square root of the rate.

Explanation of Solution

The reaction is as follows:

$\text{A+B}\to \text{AB}$

Rate law of the reaction is as follows:

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

Rate law experimentally determined by using initial rates method. The reaction is second order in reactant A. Therefore, rate is $k{\left[\text{A}\right]}^2{\left[\text{B}\right]}^n$.

Construct a table. In trial 2 concentration of reactant A is halved.

Trial	$\left[\text{A}\right]$ (M)	$\left[\text{B}\right]$ (M)	Initial rate $\left(\text{mol/L}\text{.s}\right)$
1	$1.00$	$1.00$	$1.00$
2	$0.50$	$1.00$	?

Since, order is 2 regarding reactant A, the rate is affected by the square of the concentration of the reactant A. If the concentration of reactant A is halved compared to the first trial, the new rate will be square root of the rate in trial 1. If the rate in trial 1 is r, then it will be $\sqrt{\text{r}}$ in trial 2.

Conclusion

Rates of the reaction increased with increase in the reactant concentration.