Question

A certain first-order reaction a > b is 25% complete in 42 min at 25°c. What is its rate constant?

Answer 1

Answer: Rate constant is $6.19\times 10^{-3} min^{-1}$

Explanation: For First order reaction, the expression for rate constant is:

$k= \frac{2.303}{t}log\frac{a}{a-x}$

k= rate constant

t= time required for a reaction

a= initial concentration

x= amount of reactant decomposed in time t

a-x= amount of reactant left after time t

If initial concentration (a) = 100

As 25% of the reactant has decomposed, x=25

(a-x)= (100-25) = 75

Putting in the values we get

$k= \frac{2.303}{42}log\frac{100}{75}$

$k=6.19\times 10^{-3} min^{-1}$