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

Hint: “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.
\[\begin{align}
& \text{k=}\dfrac{\text{2}\text{.303}}{\text{t}}\text{log}\left[ \dfrac{\text{concentration of the reactant}}{\text{concentration of the product}} \right] \\
& k=\dfrac{2.303}{15}\left[ \dfrac{0.8}{0.4} \right]\to (1) \\
\end{align}\]
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
\[\begin{align}
& \text{k=}\dfrac{\text{2}\text{.303}}{\text{t}}\text{log}\left[ \dfrac{\text{concentration of the reactant}}{\text{concentration of the product}} \right] \\
& k=\dfrac{2.303}{t}\left[ \dfrac{0.1}{0.025} \right]\to (2) \\
\end{align}\]
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.
\[\begin{align}
& \dfrac{2.303}{t}\left[ \dfrac{0.1}{0.025} \right]=\dfrac{2.303}{15}\left[ \dfrac{0.8}{0.4} \right] \\
 & \text{on solving the above calculation we will get} \\
 & \text{t =30min} \\
\end{align}\]
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\[{{t}_{{}^{1}/{}_{2}}}\].