Question

using the given data, calculate the rate constant of this reaction. A+B yields C+D Trial: 1,2,3 [A](M): 0.370, 0.370, 0.592 [B](M):0.340, 0.782, 0.340 Rate(M/s): 0.0210, 0.111, 0.0336 K= ? Please provide the units as well. Please show work so I can understand the problem.

Answer 1

R = k (A)^m (B)ⁿ

0.0210= k (0.370)^m (0.340 )ⁿ ------------ 1

0.111 = k (0.370)^m (0.782 )ⁿ ------------ 2

0.0336 = k (0.592)^m (0.340 )ⁿ ------------ 3

Now,

Divide eqn.-1 by eqn-2.

(0.0210) /(0.111) =( k (0.370)^m (0.340 )ⁿ ) / k (0.370)^m (0.782 )ⁿ

0.1892 == (0.4347)ⁿ

Taking log on both sides,

So, n=2.0

Now.

Divide eqn.-1 by eqn-3.

0.0210  /(0.0336) =( k (0.370)^m (0.340 )ⁿ ) / k (0.592)^m (0.340)ⁿ

0.625 ==(0.625)^m

So, m=1

overall order = m+n = 2+1 = 3

From equation-1:

0.0210 = k (0.370 )¹(0.340)²

k = 0.49

units of k in third order is M^-2 s^-1