What is the rate law of this equation?

Question

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

X + Y yields products

Trial [X] [Y] Rate

1 0.20 M 0.15 M 2.4 × 10-2 M/min

2. 0.20 M 0.30 M 4.8 × 10-2 M/min

3 0.40 M 0.30 M 19.2 × 10-2 M/min

Answer 1

Re-writing the data in a more readable form:

Trial........[X]..........[Y]..........Rate........

1..........0.2 M......0.15 M......2.4x10^-2 M/min

2..........0.2 M.....0.30 M......4.8x10^-2 M/min

3..........0.4 M.....0.30 M......19.2x10^-2 M/min

Finding the order with respect to A: compare trial 2 and 3 where [A] doubles, [B] remains the same and rate increases 4 times. Thus, the order for A is second order, or [A]²

Finding the order with respect to B: compare trial 1 and 2 where [B] doubles, [A] remains the same and the rate doubles. Thus, the order for B is first oder, or [B].

Rate Law: rate = k[A]²[B]

To determine the value of k, simply take any trial and calculate k. Thus, using trial 1 for example we have...

2.4x10^-2 M/min = k[0.2]²[0.15] and k = 2.4x10^-2 Mmin^-1/ 0.006 M³ = 4.0 M^-2min^-1

Rate = 4.0 M^-2min^-1[A]²[B] as the final rate law