Question

If rate constant of first order reaction is 1.386 × 10-² s then the half life of the reaction will be
a.100 s
b. 200 s
c. 50 s
d. 150 s​

Answer 1

Answer:

c. 50s

Explanation:

half life of first order reaction = 0.693/K, where K is rate of reaction

So, half life = 0.693/K = 0.693/1.386×10^-2 = 100/2 = 50s

hope this will help you

Answer 2

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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