Question

For a reaction $A + 2B \rightarrow C + D$, the following data were obtained

Expt. Initial concentration (moles litre–1)			Initial Rate of formation of D (moles litre–1 min–1)
S. No.	[A]	[B]
1.	0.1	0.1	6.0 × 10−3
2.	0.3	0.2	7.2 × 10−2
3.	0.3	0.4	2.88 × 10−1
4.	0.4	0.1	2.4 × 10−2

The correct rate law expression will be

• A. $\text{Rate } = \ k\lbrack A\rbrack\lbrack B\rbrack$
• B. $\text{Rate } = \ k\lbrack A\rbrack\lbrack B\rbrack^{2}$
• C. $\text{Rate } = \ k\lbrack A\rbrack^{2}\lbrack B\rbrack^{2}$
• D. $\text{Rate} = k\lbrack A\rbrack^{2}\lbrack B\rbrack$

Answer 1

Answer: (B)

$\text{Rate } = \ k\lbrack A\rbrack\lbrack B\rbrack^{2}$

Answer 2

From 1 and 4, keeping [B] constant, [A] is made 4 times, rate also becomes 4 times. Hence rate $\propto \lbrack A\rbrack$. From 2 and 3 keeping [A] constant, [B] is doubled, rate becomes 4 times. Hence rate $\propto \lbrack B\rbrack^{2}$. Overall rate law will be : rate = $k\lbrack A\rbrack\lbrack B\rbrack^{2}$