Question

If rate constant of first order reaction is 1.386 × 10–2 s–1 then the half life of the reaction will be

100 s

200 s

50 s

150 s

Answer 1

Given : rate constant of first order reaction is 1.386 × 10–2 s–1

To Find :  half life of the reaction

Solution:

first order reaction

$A= A_0e^{-kt}$

=> t = (1/k) ln(A₀/A)

A₀ = initial

A = remaining

t = time

k =  rate constant  

t will be half time if  A = A₀/2

=> A₀/A = 2

k = 1.386 * 10⁻²  per sec

=> t = (1/ 1.386 * 10⁻² ) ln(2)

=> t ≈  50

Hence half life of the reaction will be 50  sec

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Answer 2

PLEASE NOTE:

• For school exams, please follow the above answer by @Amitnrw sir.

• Use my answer only for MCQs, it's a fast method !

Half life in 1st order reaction is :

$\boxed{ t_{ \frac{1}{2} } = \dfrac{0.693}{k} }$

• 'k' is rate constant.

$\implies t_{ \frac{1}{2} } = \dfrac{0.693}{1.386 \times {10}^{ - 2} }$

$\implies t_{ \frac{1}{2} } = \dfrac{0.693}{1.386 } \times 100$

$\implies t_{ \frac{1}{2} } = 0.5 \times 100$

$\implies t_{ \frac{1}{2} } = 50 \: sec$

So, half life of the reaction is 50 Seconds.