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The reaction is second order in A. When [A]₀ = 0.100 M, the reaction is 20.0% complete in 35.9 minutes. Calculate the value of the rate constant (in L/min·mol).

A) 6.96 x 10⁻²
B) 5.57 x 10⁻⁴
C) 1.57
D) 1.11
E) none of these

Answer :

To find the rate constant for a second-order reaction, we use the rate law expression. For [A]0 = 0.100 M and 20% completion in 35.9 minutes, the rate constant is found to be 6.96 x 10^-2 L/min·mol, which matches answer choice A.

To calculate the rate constant for a reaction that is second order in A, we need to use the second-order rate law expression: 1/[A] = kt + 1/[A]0

Where [A] is the concentration of reactant A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time.

Given that [A]0 = 0.100 M and the reaction is 20.0% complete in 35.9 minutes, the concentration [A] can be found with the following: [A] = [A]0 - 0.20 * [A]0. Thus:

[A] = 0.100 M - 0.20 * 0.100 M = 0.100 M - 0.020 M = 0.080 M

Now we can plug in the values to the second-order rate law expression and solve for the rate constant k:

1/0.080 M = k * 35.9 min + 1/0.100 M

12.5 L/mol = k * 35.9 min + 10 L/mol

2.5 L/mol = k * 35.9 min

k = (2.5 L/mol) / (35.9 min)

k = 0.0696 L/min·mol

The rate constant is therefore 6.96 x 10^-2 L/min·mol, which corresponds to answer choice A.

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