Question

In a first-order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is:
A. 30 min
B. 15 min
C. 7.5min
D. 60min

Answer 1

“A first-order reaction is a reaction that proceeds at a rate that depends linearly on only concentration of one reactant”. The formula to calculate the rate constant of a first order reaction is
$\text{k=}\dfrac{\text{2}\text{.303}}{\text{t}}\text{log}\left[ \dfrac{\text{concentration of the reactant}}{\text{concentration of the product}} \right]$
K = rate constant of the first order reaction
t = half-life of the reaction

Complete step by step solution:
In the question, it is mentioned that the order of the reaction is first order.
In the question it is given that the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 min (half-life of the reaction).
$\text{k=}\dfrac{\text{2}\text{.303}}{\text{t}}\text{log}\left[ \dfrac{\text{concentration of the reactant}}{\text{concentration of the product}} \right]$
k = rate constant of the first order reaction
t = half-life of the reaction
Now substitute all the known values in the above reaction.

& \text{k=}\dfrac{\text{2}\text{.303}}{\text{t}}\text{log}\left[ \dfrac{\text{concentration of the reactant}}{\text{concentration of the product}} \right] \\\ & k=\dfrac{2.303}{15}\left[ \dfrac{0.8}{0.4} \right]\to (1) \\\ \end{aligned}$$ Coming to the second case the concentration of the reactant decreases from 0.1 M to 0.025 M and it is also a first order reaction. Then $$\begin{aligned} & \text{k=}\dfrac{\text{2}\text{.303}}{\text{t}}\text{log}\left[ \dfrac{\text{concentration of the reactant}}{\text{concentration of the product}} \right] \\\ & k=\dfrac{2.303}{t}\left[ \dfrac{0.1}{0.025} \right]\to (2) \\\ \end{aligned}$$ For all first order reactions if reactants are same then the rate constants are also same. Then we can substitute equation (2) equation (1) to get half-life (t) of second reaction. $$\begin{aligned} & \dfrac{2.303}{t}\left[ \dfrac{0.1}{0.025} \right]=\dfrac{2.303}{15}\left[ \dfrac{0.8}{0.4} \right] \\\ & \text{on solving the above calculation we will get} \\\ & \text{t =30min} \\\ \end{aligned}$$ Therefore time taken for the concentration to change from 0.1 M to 0.025 M is 30 min. **So, the correct option is A, 30 min.** **Note:** Half-life in a chemical reaction is the time required for a quantity or concentration of a chemical to reduce to half of its initial quantity or concentration. The half-life of a chemical reaction is going to be denoted with a symbol$${{t}_{{}^{1}/{}_{2}}}$$.