SAT Chemistry

Category - Multiple Choice

Given the following experimental values, for the reaction A + B --> C, figure out the rate constant (k).

Initial values: [A] = 0.0100M; [B] = 0.0100M; Rate = 5.00 * 10-6 M/s
Experiment 1: [A] = .0200M; [B] = 0.100M; Rate = 2.50 * 10-5 M/s
Experiment 2: [A]= 0.0100M; [B]=0.0200M; Rate = 5.00 * 10-6 M/s
  1. k = 5.00 * 10-2 M*s
  2. k = 5.00 * 10-2/(M*s)
  3. k = 5.00 * 10-2/(M2*s)
  4. k = 5.00 * 10-3 M*s
  5. k = 5.00 * 10-3/(m*s)
Explanation
Answer: B. Solution is as follows:

When [A] doubles, the rate is 4 times as large (2.50 * 10-5/5.00 *10-6 = 4). Thus, we have [2A] = 4r. Therefore, we have a second order reaction with respect to A.

When B doubles, the reaction rate doesn’t change, meaning we have a zero order with respect to B.

Thus r=k[A]2

Inputting the initial values:

5.0 *10-6 M/s = k[0.0100M]2
solving for k yields 5.00 *10-2 with units of 1/(M*s)
Kinetic control vs. thermodynamic control
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