Question

At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reaction be if the concentration of H2 were halved

Answer 1

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

$rate=k[H_2]^2[NH_3]$

At a certain concentration of $H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]$

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

$0.120=k[H_2]^2[NH_3]$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[NH_3]$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}$

$\frac{R}{0.120}=\frac{1}{4}$

$R=0.03M/s$

Therefore, the initial rate of the reaction will be, 0.03 M/s