Q126 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Chemistry
A first order reaction is 50% completed in 10 minutes. Find the percentage of reactant reacting in 20 minutes.
- A$25\%$
- B$100\%$
- Correct answerC$75\%$
- D$20\%$
Solution
For a first order reaction, the half-life $t_{1/2}$ is given as $10$ minutes.
The amount of reactant remaining after $n$ half-lives is given by $N = N_0 \left(\dfrac{1}{2}\right)^n$.
Time $t = 20$ minutes corresponds to $n = \dfrac{20}{10} = 2$ half-lives.
Amount of reactant remaining after $20$ minutes is $N = N_0 \left(\dfrac{1}{2}\right)^2 = \dfrac{N_0}{4} = 25\%$ of $N_0$.
The percentage of reactant that has reacted is $100\% - 25\% = 75\%$.
Answer: $75\%$
The amount of reactant remaining after $n$ half-lives is given by $N = N_0 \left(\dfrac{1}{2}\right)^n$.
Time $t = 20$ minutes corresponds to $n = \dfrac{20}{10} = 2$ half-lives.
Amount of reactant remaining after $20$ minutes is $N = N_0 \left(\dfrac{1}{2}\right)^2 = \dfrac{N_0}{4} = 25\%$ of $N_0$.
The percentage of reactant that has reacted is $100\% - 25\% = 75\%$.
Answer: $75\%$
Hence, the correct answer is option: 'C'.