Question

Trial	[A] (M)	[B] (M)	Rate (M/s)
1	0.400	0.350	0.0196
2	0.400	0.700	0.0196
3	0.800	0.350	0.0784

Using the data in the table, determine the rate constant of the reaction and select the appropriate units.

A+2B⟶C+D

k=? units=?

Answer 1

Here the reaction is given , A + 2B ⟶ C + D .

We know the Rate law of a chemical reaction can be written in the form,

Rate = k x [A]^a x [B]^b

Where , k = rate constant, A and B are reactants , a and b are order of reaction with respect to A and B.

Hence here,

Rate = k x [A]^a x [B]^b

According to the given information,

When the concentration of A is kept constant and concentration of B changed from 0.350 M to 0.700 M, the rate of reaction do not changed, it is 0.0196 M/s.

That is, when only the concentration of B changed, the rate of reaction is independent of concentration of B. It clearly indicate that the reaction is zero order with respect to B.

Similarly, when the concentration of B is kept constant and concentration of A changed from 0.400 M to 0.800 M, the rate of reaction changed from 0.0196 M/s to 0.0784 M/s.

0.800 / 0.400 = 2

0.0784 / 0.0196 = 4

That is when only the concentration of A got 2 times, the rate of reaction got 4 times. It clearly indicate that the reaction is second order with respect to A .

Hence a=2 , b=0

( Note : The total order of reaction is 2. )

Hence The rate law can be written as,

Rate = k x [A]² x [B]⁰

Rate = k x [A]²

To find out the value of k, we can substitute values for 1 case in the above rate law,

Given,

[A] = 0.400 M

Rate = 0.0196 M/s

Rate = k x [A]²

Substituting,

0.0196 = k x 0.400²

k = 0.0196 / 0.400² = 0.1225 M^-1 s^-1 = 0.123 M^-1 s^-1

Please note that the value of k has 3 significant figures since the operands have a minimum of 3 significant figures.

Unit of K is M^-1 s^-1​​​​​​​ .

Best Wishes.