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[I_2]^2[/tex]
At a certain concentration of [tex]H_2[/tex] and [tex]I_2[/tex], 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, [tex]1.0\times 10^4M/s[/tex]
Explanation :
Rate law expression for the reaction:
[tex]rate=k[H_2]^2[I_2]^2[/tex]
As we are given that:
Initial rate = 4.0 × 10⁴ M/s
Expression for rate law for first observation:
[tex]4.0\times 10^4=k[H_2]^2[I_2]^2[/tex] ....(1)
Expression for rate law for second observation:
[tex]R=k(\frac{[H_2]}{2})^2[I_2]^2[/tex] ....(2)
Dividing 2 by 1, we get:
[tex]\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}[/tex]
[tex]\frac{R}{4.0\times 10^4}=\frac{1}{4}[/tex]
[tex]R=1.0\times 10^4M/s[/tex]
Therefore, the initial rate of the reaction will be, [tex]1.0\times 10^4M/s[/tex]
