Question

Consider the reaction, \[\ce{A + B -> C + D}\]

The initial rate for different initial concentrations of the reactants are given below:

S. No.	Initial concentration (mol L−1)		Initial rate (mol L−1 s−1)
	A	B
(i)	1.0	1.0	2.0 × 10−3
(ii)	2.0	1.0	4.0 × 10−3
(iii)	4.0	1.0	8.0 × 10−3
(iv)	1.0	2.0	2.0 × 10−3
(v)	1.0	4.0	2.0 × 10−3

1. What are the orders with respect to A and B?
2. What is the overall order?
3. Write the rate law equation.
4. Calculate the rate constant.
5. Suggest a possible mechanism.

Answer 1

Given: r ∝ [A]^x [B]^y

a. Orders wrt A and B:

Compare (ii)/(i):

`(4.0 xx 10^-3)/(2.0 xx 10^-3) = (2/1)^x`

⇒ 2 = 2x

⇒ x = 1.

Compare (iv)/(i):

`(2.0 xx 10^-3)/(2.0 xx 10^-3) = (1/1)^x (2/1)^y`

⇒ 1 = 2y

⇒ y = 0.

∴ Order in A = 1, Order in B = 0

b. Overall order:

Overall order = x + y 

= 1 + 0

= 1

c. Rate law:

r = k [A]

d. Rate constant k:

Using row (i): k = `r/([A])`

= `(2.0 xx 10^-3)/1.0`

= 2.0 × 10⁻³ s⁻¹

e. Possible mechanism:

We know that, r = k[A]

A slow unimolecular step is followed by a fast step with B:

\[\ce{A ->[Slow] I}\];

\[\ce{I + B ->[Fast] C + D}\]