The rate of a certain reaction is given by the following rate law:
rate=k[N2]^3[O3]
Use this information to answer the question below.
At a certain concentration of N2 and O3, the initial rate of reaction is 0.780 M / s. What would the initial rate of the reaction be if the concentration of N2 were halved? Round your answer to significant digits.
If the rate law is given by rate = k[N2]^3[O3], we can assume that the rate constant k and the concentration of O3 are constant.
If the concentration of N2 is halved, it means that [N2] becomes half of its initial concentration. Let's call the initial concentration of N2 as [N2]1 and the halved concentration as [N2]2 = [N2]1 / 2.
Using the rate law, we can compare the initial rates of the reaction:
rate1 = k[N2]1^3[O3]
rate2 = k[N2]2^3[O3]
Substituting [N2]2 = [N2]1 / 2 into rate2:
rate2 = k([N2]1 / 2)^3[O3]
= (k[N2]1^3[O3]) / 8
Since the concentration of O3 and k are constant, we can see that rate2 is 1/8 of rate1.
Therefore, if the concentration of N2 is halved, the initial rate of the reaction would be 0.780 M/s / 8 = 0.098 M/s.