Question

The following results were obtained during kinetic studies of the reaction:

2A + B → 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 × 10-3
(II)	0.10	0.25	6.93 × 10-3
(III)	0.20	0.30	1.386 × 10-2

 

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

1. 10
2. 5
3. 100
4. 1

Answer 1

Correct Answer - Option 1 : 10

Calculation:

Here A and B are reactant,

Let the rate expression is r ∝ [A]^a [B]^b

Here, r is rate of expression

By taking from experiment 1,

\(\frac{{{r_2}}}{{{r_1}}} = {\left( {\frac{{0.1}}{{0.1}}} \right)^a} \times {\left( {\frac{{0.25}}{{0.25}}} \right)^b}\)

\(\frac{{6.93{\rm{\;}} \times {{10}^{ - 2}}}}{{6.93{\rm{\;}} \times {{10}^{ - 2}}}} = 1 \times {\left( {\frac{5}{4}} \right)^b}\)

\(1 = {\left( {\frac{5}{4}} \right)^b} \Rightarrow {\left( {\frac{5}{4}} \right)^0} = {\left( {\frac{5}{4}} \right)^b} \Rightarrow b = 0\)

From the experiment 2, \(\frac{{{r_3}}}{{{r_1}}} = {\left( {\frac{{0.2}}{{0.1}}} \right)^a} \times {\left( {\frac{{0.30}}{{0.1}}} \right)^b}\)

\(\frac{{1.386{\rm{\;}} \times {{10}^{ - 2}}}}{{0.6.93{\rm{\;}} \times {{10}^{ - 2}}}} = {2^a} \times {\left( {1.5} \right)^a}\)

2 = 2^a × 1 ⇒ 2¹ = 2^a ⇒ a = 1

So r ∝ [A]¹ [B]⁰ ⇒ r ∝ [A]

 Order of the reaction (n) = 1

Now, let for the 1^st experiment

r₁ = k.[A]

\(K = \frac{{{r_1}}}{{\left[ A \right]}}\)

\(= \frac{{6.93{\rm{\;}} \times {{10}^{ - 2}}}}{{0.1}}\)

 = 6.93 × 10^-2 s^-1

\({t_{50}} = {\rm{\;}}\frac{{0.6.93}}{{6.93{\rm{\;}} \times {{10}^{ - 2}}}} = 10\;s\)