Question

The results given in the below table were obtained during kinetic studies of the following reaction
\(2A + B \to C + D\)

Experiment	[A]/mol L-1	[B]/mol L-1	Initial rate/mol L-1 min-1
I	0.1	0.1	\(6.00 \times 10^{-3}\)
II	0.1	0.2	\(2.40 \times 10^{-2}\)
III	0.2	0.1	\(1.20 \times 10^{-2}\)
IV	X	0.2	\(7.20 \times 10^{-2}\)
V	0.3	Y	\(2.88 \times 10^{-1}\)

X and Y in the given table are respectively :

• A. 0.3, 0.4
• B. 0.4, 0.3
• C. 0.4, 0.4
• D. 0.3, 0.3

Hint:

Find the orders from experiments I, II, III, then the rate constant, and use it for IV and V.

Answer 1

The Correct Option is A

Let rate \(= k[A]^a[B]^b\).
I and II: [B] doubles, rate \(\times 4\): \(2^b = 4\), so \(b = 2\).
I and III: [A] doubles, rate \(\times 2\): \(2^a = 2\), so \(a = 1\).
\(k = \frac{6.00\times10^{-3}}{(0.1)(0.1)^2} = 6\ \mathrm{L^2\,mol^{-2}\,min^{-1}}\).
Exp IV: \(7.20\times10^{-2} = 6\times X\times(0.2)^2\), so \(X = \frac{0.072}{0.24} = 0.3\).
Exp V: \(2.88\times10^{-1} = 6\times0.3\times Y^2\), so \(Y^2 = \frac{0.288}{1.8} = 0.16\), \(Y = 0.4\).
Answer: (1) 0.3, 0.4