Question

Using the data in the table, calculate the rate constant of this reaction. 

Trial	[ A(M)	[B(M)	Rate (M//s)
1	0.300	0.390	0.0176
2	0.300	0.897	0.0931
3	0.480	0.390	0.0282

A+B → C+D

Answer 1

k = 0.386

Units : M^-2.s^-1

Explanation

For trial 1 and 2, concentration of reactant A remains constant.

order of reaction with respect to reactant B = log(R₂ / R₁) / log([B]₂ / [B]₁)

order of reaction with respect to reactant B = log(0.0931 M/s / 0.0176 M/s) / log(0.897 M / 0.390 M)

order of reaction with respect to reactant B = 2

For trial 1 and 3, concentration of reactant B remains constant.

order of reaction with respect to reactant A = log(R₃ / R₁) / log([A]₃ / [A]₁)

order of reaction with respect to reactant A = log(0.0282 M/s / 0.0176 M/s) / log(0.480 M / 0.300 M)

order of reaction with respect to reactant A = 1

rate law is : rate = k[A][B]²

k = rate / [A][B]²

For trial 1,

k = 0.0176 M/s / [(0.300 M) * (0.390 M)²]

k = 0.386 M^-2.s^-1

For trial 2,

k = 0.0931 M/s / [(0.300 M) * (0.897 M)²]

k = 0.3857 M^-2.s^-1

For trial 3,

k = 0.0282 M/s / [(0.480 M) * (0.390 M)²]

k = 0.3863 M^-2.s^-1

Average k = 0.386 M^-2.s^-1