Question
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.
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.
Answers
GPT 3.5
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.
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.