Question

At a certain concentration of N2 and O3, the initial rate of reaction is 0.770 M / s. What would the initial rate of the reaction be if the concentration of N2 were halved? Be sure your answer has the correct number of significant digits.

Answer 1

The question is incomplete, here is the complete question:

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

$rate=k[N_2]^2[H_2]^2$

At a certain concentration of $N_2$ and $H_2$, the initial rate of reaction is 0.770 M/s. What would the initial rate of the reaction be if the concentration of $N_2$ were halved? Be sure your answer has the correct number of significant digits.

Answer : The initial rate of the reaction will be 0.192 M/s

Explanation :

Rate law expression for the reaction:

$rate=k[N_2]^2[H_2]^2$

As we are given that:

Initial rate = 0.770 M/s

Expression for rate law for first observation:

$0.770=k[N_2]^2[H_2]^2$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[N_2]}{2})^2[H_2]^2$ ....(2)

Dividing 1 by 2, we get:

$\frac{R}{0.770}=\frac{k(\frac{[N_2]}{2})^2[H_2]^2}{k[N_2]^2[H_2]^2}$

$\frac{R}{0.770}=\frac{1}{4}$

$R=0.192M/s$

Therefore, the initial rate of the reaction will be 0.192 M/s