Question

The time required for $60\%$ completion of a first order reaction is $50$ min. The time required for $93.6\%$ completion of the same reaction will be

• A. 100 min
• B. 83.8 min
• C. 50 min
• D. 150 min

Answer 1

The Correct Option is D

For a first order reaction the, the rate constant is given by $k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$
Given, at $50 \,min , 60 \%$ of the reaction is completed
$\therefore k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$ $=\frac{2.303}{50} \log \frac{100}{40}$ $=\frac{2.303}{50} \times 0.397$
So, when $93.6 \%$ of the reaction is completed,
$\Rightarrow \frac{2.303}{50} \times 0.397$ $=\frac{2.303}{t} \log \frac{100}{6.4}$ $\Rightarrow \frac{2.303}{50} \times 0.397$ $=\frac{2.303}{t} \times 1.19$ $\Rightarrow t \approx 150\, min$

Answer 2

For a first-order reaction, the rate law can be expressed as: \\ \[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \] Where, \([A_0]\) is the initial concentration, \([A]\) is the concentration after time \(t\), \(k\) is the rate constant, and \(t\) is the time taken. The equation can be written as: \[ t = \frac{1}{k} \ln \left( \frac{[A_0]}{[A]} \right) \] For the first case, where 60\% of the reaction is complete: \[ [A] = 0.40 [A_0] \quad \text{(because 60\% is completed, so 40\% remains)} \] Substitute into the equation to find \(k\): \[ 50 = \frac{1}{k} \ln \left( \frac{1}{0.40} \right) \] \[ 50 = \frac{1}{k} \ln(2.5) \] \[ 50 = \frac{1}{k} \times 0.9163 \quad \Rightarrow k = \frac{0.9163}{50} = 0.01833 \, \text{min}^{-1} \] Now, for the second case, where 93.6\% of the reaction is complete: \[ [A] = 0.064 [A_0] \quad \text{(because 93.6\% is completed, so 6.4\% remains)} \] Substitute into the equation to find \(t\): \[ t = \frac{1}{0.01833} \ln \left( \frac{1}{0.064} \right) \] \[ t = \frac{1}{0.01833} \times 2.7726 = 151.56 \, \text{min} \] Thus, the closest correct option is 150 min.