For the reaction, 4 A (g) + 3 B (g) = > 2 C (g), the following data were obtained at constant temperature.

Experiment Initial A, mol/L Initial B, mol/L Initial Rate, M/min

1 0.200 0.150 5.00

2 0.400 0.150 10.0

3 0.200 0.300 20.0

4 0.400 0.300 40.0

1. Determine the value of k for the reaction.

2. Which of the following is the correct rate law for the reaction?1. Rate = k[A]2[B]22. Rate = k[A][B]3. Rate = k[A]2[B]4. Rate = k[A][B]2

The k value for the reaction is k=5.56*10^3 M^-3/min and the rate law will be in order 3.

Explanation:

Finding the K value for the reaction

Write the rate equation,

Rate=K[A]*[A] [B]*[B]

Taking the rate values from the equation no 1 data

let K=5.00 M/min A=0.200 B=0.150

Rate=K[A]*[A] [B]*[B]

5.00 M/min = k (0.200) 2 (0.150) 2 = 0.0009k

k = 5.00/0.0009

k=5556

k=5.56*10^3 M^-3/min

Then by solving the rate law we will get,

The Rate law will be k[A]2[B]4.