The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
[tex]rate=k[H_2]^2[NH_3][/tex]
At a certain concentration of [tex]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/s
Explanation :
Rate law expression for the reaction:
[tex]rate=k[H_2]^2[NH_3][/tex]
As we are given that:
Initial rate = 0.120 M/s
Expression for rate law for first observation:
[tex]0.120=k[H_2]^2[NH_3][/tex] ....(1)
Expression for rate law for second observation:
[tex]R=k(\frac{[H_2]}{2})^2[NH_3][/tex] ....(2)
Dividing 2 by 1, we get:
[tex]\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}[/tex]
[tex]\frac{R}{0.120}=\frac{1}{4}[/tex]
[tex]R=0.03M/s[/tex]
Therefore, the initial rate of the reaction will be, 0.03 M/s
