Quest question on Kinetics

Question

A chemical reaction is expressed by the balanced chemical equation A + 2 B −→ C

Consider the data Initial Initial Initial [A] [B] rate M M M/min

1 0.18 0.18 0.00142884

2 0.18 0.36 0.00571536

3 0.36 0.36 0.0114307

Find the rate law for the reaction. 1. R = k[A] [B]2 2. R = k[A]2 [B] 3. R = k[A] [B]

Part 2

If the initial concentrations of both A and B are 0.41 M, at what initial rate is C formed? Answer in units of M/min.

Answer 1

Your table isn't formatted properly, so I'm sort of guessing at the actual data being presented.

A + 2B ==> C

Initial A Initial B rate M/min

0.18..........0.18.........0.00142884 trial 1

0.18...........0.36........0.00571536 trial 2

0.36...........0.36........0.0114307 trial 3

Comparing trial 1 and 2, [B] doubles while [A] remains constant, and rate increases by 4 times. This tells us the reaction is 2nd order in B.

Comparing trial 1 and 3, [A] doubles and [B] doubles and rate increases by 8 times. Since the reaction is 2nd order in B, this tells us that the reaction is 1st order in A

Rate = k[A][B]²

Part 2:

Using any trial, we can find the value of k. I will use trial 3:

0.0114307 M/min = k(0.36 M][0.36 M]² = 0.004666 M³

k = 0.0114307 M/min / 0.04666 M³

k = 0.24498 M^-2 min^-1

Now, to find the rate of formation of C, when both A and B = 0.41 M, we have...

rate = 0.24498 M^-2 min^-1 (0.41 M)(0.41 M)² = 0.24498 M^-2 min^-1 x 0.068921 M³

rate = 0.016884 M/min