Question

For a particular first-order reaction, it takes 48 minutes for the concentration of the reactant to decrease to 25% of its initial value. What is the value for rate constant (in s -1) for the reaction

Answer 1

Answer: The value for rate constant for a reaction is $4.81\times 10^{-4} s^{-1}$

Explanation:

The integrated rate law equation for first-order kinetics:

$k=\frac{2.303}{t}\log \frac{a}{a-x}$ ......(1)

Let the initial concentration of reactant be 100 g

Given values:

a = initial concentration of reactant = 100 g

a - x = concentration of reactant left after time 't' = 25 % of a = 25 g

t = time period = 48 min = 2880 s       (Conversion factor: 1 min = 60 s)

Putting values in equation 1:

$k=\frac{2.303}{2880s}\log (\frac{100}{25})\\\\k=4.81\times 10^{-4} s^{-1}$

Hence, the value for rate constant for a reaction is $4.81\times 10^{-4} s^{-1}$