Half-life period of a first order reaction is 10 min. Starting with initial concentration 12 M, the rate after 20 min is

Question

Half-life period of a first order reaction is 10 min. Starting with initial concentration 12 M, the rate after 20 min is (A) 0.0693 M min-1 (B) 0.693 × 3 M min -1 (C) 0.0693 × 3 M min-1 (D) 0.0693 × 4 M min-1

Tardigrade Admin · Accepted Answer

Key concept As we know, rate constant remains same throughout the reaction. For first order reaction, half-life of the reactant does not depend upon the its concentration. Hence, we can easily calculate the value of rate constant with the help of provided half-life value by using the formula:

t_{1/2} = \frac{0.693}{k}

. Now, we can calculate the

concentration of reactant after

20 min

and then rate of reaction.

Given,

t_{1/2} = 10 min

S in t_{1/2} = \frac{0.693}{k}

(k = rate constant) ;

k = \frac{0.693}{10} = 0.0693

Stage	Time	Concentration
Intial	t=0	12 M
	$t_{1/2} = 10 min$	6 M
	$t = 20 min$	3 M

Since, it is a first order reaction.

Rate

= k [A] = 0.0693 \times 3 M mi n^{- 1}

Q. Half-life period of a first order reaction is $10 min$ . Starting with initial concentration $12 M$ , the rate after $20 min$ is

$0.0693 M mi n^{- 1}$

$0.693 \times 3 M mi n^{- 1}$

$0.0693 \times 3 M mi n^{- 1}$

$0.0693 \times 4 M mi n^{- 1}$

Q. Half-life period of a first order reaction is 10min. Starting with initial concentration 12M, the rate after 20min is

0.0693Mmin−1

0.693×3Mmin−1

0.0693×3Mmin−1

0.0693×4Mmin−1

Q. Half-life period of a first order reaction is $10 min$ . Starting with initial concentration $12 M$ , the rate after $20 min$ is

$0.0693 M mi n^{- 1}$

$0.693 \times 3 M mi n^{- 1}$

$0.0693 \times 3 M mi n^{- 1}$

$0.0693 \times 4 M mi n^{- 1}$