Experiment Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L. s 1 0.0500 0.0500 0.0100 6.25 x 10-3 2 0.100 0.0500 0.0100 1.25 x 10-2 3 0.100 0.100 0.0100 2.50 x 10-2 4 0.0500 0.0500 0.0200 1.25 x 10-2What is the rate law for the reaction?a) k[A][B][C] b) k [A]2[B][C]

k[A][B][C]

Explanation:

Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L.s

1      0.0500       0.0500        0.0100       6.25 x 10-3

2      0.100         0.0500        0.0100        1.25 x 10-2

3      0.100         0.100            0.0100       2.50 x 10-2

4      0.0500      0.0500         0.0200      1.25 x 10-2

Comparing equations 1 and 2, the reaction is first order with respect to A. This is because the concentration of A doubles, while the concentration of B and C remained constant leading to a double of the rate of reaction.

Comparing equations 2 and 3, the reaction is first order with respect to B. This is because the concentration of B doubles, while the concentration of A and C remained constant leading to a double of the rate of reaction.

Comparing equations 1 and 4, the reaction is first order with respect to C. This is because the concentration of C doubles, while the concentration of A and B remained constant leading to a double of the rate of reaction.

This means our rate law = k[A][B][C];

That is first order with respect to A, B and C

the answer is symmetrical : )