Which of the following correctly matches the integrated rate law for a simple second-order reaction (where the rate depends on the concentration of a single reactant A)?

• A. [A] = [A]_0 - kt
• B. [A] = [A]_0 e^{-kt}
• C. 1/[A] = 1/[A]_0 + kt
• D. ln[A] = ln[A]_0 - kt

Answer: (C)

Verified step by step guidance

1. Identify the order of the reaction and the corresponding integrated rate law. For a second-order reaction where the rate depends on a single reactant A, the rate law is given by: \[ rate = k[A]^2 \].
2. Recall the integrated rate law for a second-order reaction. It relates the concentration of A at time t, \[[A]\], to the initial concentration \[[A]_0\] and the rate constant k as: \[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \].
3. Compare the given options with the known integrated rate laws for zero, first, and second-order reactions: - Zero order: \[ [A] = [A]_0 - kt \] - First order: \[ \ln[A] = \ln[A]_0 - kt \] - Second order: \[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \].
4. Understand why the other options do not fit a second-order reaction: - \[ [A] = [A]_0 - kt \] corresponds to zero order because concentration decreases linearly with time. - \[ [A] = [A]_0 e^{-kt} \] and \[ \ln[A] = \ln[A]_0 - kt \] correspond to first order because the concentration decreases exponentially or logarithmically with time.
5. Conclude that the correct integrated rate law for a simple second-order reaction with one reactant A is: \[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \], which matches the third option.