Question

At a certain concentration of H2 and I2, the initial rate of reaction is 4.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were halved? Round your answer to

Answer 1

The question is incomplete, here is the complete question:

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

$rate=k[H_2]^2[I_2]^2$

At a certain concentration of $H_2$ and $I_2$, the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of

Answer : The initial rate of the reaction will be, $1.0\times 10^4M/s$  

Explanation :

Rate law expression for the reaction:

$rate=k[H_2]^2[I_2]^2$

As we are given that:

Initial rate = 4.0 × 10⁴ M/s

Expression for rate law for first observation:

$4.0\times 10^4=k[H_2]^2[I_2]^2$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[I_2]^2$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}$

$\frac{R}{4.0\times 10^4}=\frac{1}{4}$

$R=1.0\times 10^4M/s$

Therefore, the initial rate of the reaction will be, $1.0\times 10^4M/s$