Respuesta :
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
[tex]rate=k[N_2]^2[H_2]^2[/tex]
At a certain concentration of [tex]N_2[/tex] and [tex]H_2[/tex], the initial rate of reaction is 0.770 M/s. What would the initial rate of the reaction be if the concentration of [tex]N_2[/tex] 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:
[tex]rate=k[N_2]^2[H_2]^2[/tex]
As we are given that:
Initial rate = 0.770 M/s
Expression for rate law for first observation:
[tex]0.770=k[N_2]^2[H_2]^2[/tex] ....(1)
Expression for rate law for second observation:
[tex]R=k(\frac{[N_2]}{2})^2[H_2]^2[/tex] ....(2)
Dividing 1 by 2, we get:
[tex]\frac{R}{0.770}=\frac{k(\frac{[N_2]}{2})^2[H_2]^2}{k[N_2]^2[H_2]^2}[/tex]
[tex]\frac{R}{0.770}=\frac{1}{4}[/tex]
[tex]R=0.192M/s[/tex]
Therefore, the initial rate of the reaction will be 0.192 M/s
