Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) 75% of a first order reaction is completed in 32 minutes. 50% of the reaction would have been completed in ______ minutes.

Hint:

For first-order reactions, the time to reach a certain percentage completion can be calculated using \( t = \frac{\ln([A]_0/[A])}{k} \). Half-life \(t_{1/2}\) is independent of initial concentration.

Answer 1

Correct Answer: 16

Step 1: For a first-order reaction, the time \(t\) required to complete a fraction of the reaction is given by: \[ \ln \frac{[A]_0}{[A]} = k t \] where \(k\) is the rate constant, \([A]_0\) is the initial concentration, and \([A]\) is the concentration at time \(t\).
Step 2: Let \(t_{75%} = 32 \, \mathrm{min}\). For 75% completion, \([A] = 0.25[A]_0\): \[ \ln \frac{[A]_0}{0.25[A]_0} = k \cdot 32 \quad \Rightarrow \quad \ln 4 = k \cdot 32 \]
Step 3: Solve for the rate constant \(k\): \[ k = \frac{\ln 4}{32} = \frac{1.3863}{32} \approx 0.04332 \, \mathrm{min^{-1}} \]
Step 4: For 50% completion, \([A] = 0.5[A]_0\): \[ \ln \frac{[A]_0}{0.5[A]_0} = k t_{50%} \quad \Rightarrow \quad \ln 2 = k t_{50%} \]
Step 5: Solve for \(t_{50%}\): \[ t_{50%} = \frac{\ln 2}{k} = \frac{0.6931}{0.04332} \approx 16.00 \, \mathrm{min} \]