Trial # Initial A (M) Initial B (M) Initial Rate (M/s)1 .350 .200 6.0 x 10-62 .700 .200 2.4 x 10-53 .350 .400 2.4 x 10-5Given the table above determine k.Group of answer choices2.45 x 10-4 M-2s-14.90 x 10-5 M-1s-11.5 x 10-4 M-1s-11.22 x 10-3 M-3s-1
Question
Trial # Initial A (M) Initial B (M) Initial Rate (M/s)1 .350 .200 6.0 x 10-62 .700 .200 2.4 x 10-53 .350 .400 2.4 x 10-5Given the table above determine k.Group of answer choices2.45 x 10-4 M-2s-14.90 x 10-5 M-1s-11.5 x 10-4 M-1s-11.22 x 10-3 M-3s-1
Solution
The rate law for a reaction can be determined from the initial rates of reaction. The rate law has the form:
Rate = k[A]^m[B]^n
where:
- Rate is the initial rate of the reaction,
- k is the rate constant,
- [A] and [B] are the concentrations of the reactants,
- m and n are the orders of the reaction with respect to the reactants.
From the given data, we can see that when the concentration of A is doubled (from trial 1 to trial 2), the rate of the reaction is quadrupled. This suggests that the reaction is second order with respect to A (m = 2).
Similarly, when the concentration of B is doubled (from trial 1 to trial 3), the rate of the reaction is also quadrupled. This suggests that the reaction is second order with respect to B (n = 2).
Therefore, the rate law for the reaction is:
Rate = k[A]^2[B]^2
We can solve for k using the data from any of the trials. Let's use trial 1:
6.0 x 10^-6 M/s = k(0.350 M)^2(0.200 M)^2
Solving for k gives:
k = 6.0 x 10^-6 M/s / (0.350 M)^2(0.200 M)^2 = 1.22 x 10^-3 M^-3s^-1
So, the rate constant k for this reaction is 1.22 x 10^-3 M^-3s^-1.
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