Question

The following results were obtained during kinetic studies of the reaction ;
\(2A + B \rightarrow \text{Products}\)

Experiment	[A] (in mol L-1)	[B] (in mol L-1)	Initial Rate of reaction (in mol L-1 min-1)
I	0.10	0.20	\(6.93\times10^{-3}\)
II	0.10	0.25	\(6.93\times10^{-3}\)
III	0.20	0.30	\(1.386\times10^{-2}\)

The time (in minutes) required to consume half of A is :

• A. 5
• B. 10
• C. 1
• D. 100

Hint:

Find the orders from the data (zero order in B, first order in A), get k from rate/[A], then use \(t_{1/2}=0.693/k\).

Answer 1

The Correct Option is B

Rate = k[A]^x[B]^y.
Exp I and II: [A] constant, [B] changes 0.20 to 0.25 but the rate is unchanged, so y = 0.
Exp I and III: [A] doubles (0.10 to 0.20) and the rate doubles (\(6.93\times10^{-3}\) to \(1.386\times10^{-2}\)), so x = 1.
Rate = k[A]: \(k=\dfrac{6.93\times10^{-3}}{0.10}=6.93\times10^{-2}\ \mathrm{min^{-1}}\).
First-order half-life: \(t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{6.93\times10^{-2}}=10\) min.
Answer: (2) 10