Question

The first order reaction \( A(g) \rightarrow B(g) + 2C(g) \) occurs at 25\(^\circ\)C. After 24 minutes the ratio of the concentration of products to the concentration of the reactant is 1:3. What is the half-life of the reaction (in min)? (log 1.11 = 0.046)

• A. 150.5
• B. 142.2
• C. 157.8
• D. 15.78

Hint:

For first-order reactions, the half-life is independent of the initial concentration and depends only on the rate constant. Keep in mind the logarithmic relationship when calculating changes in concentration.

Answer 1

The Correct Option is C

For a first-order reaction, the equation for the change in concentration over time is: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] where: - \([A]_0\) is the initial concentration, - \([A]\) is the concentration after time \( t \), - \( k \) is the rate constant, - \( t \) is the time elapsed. We are given that after 24 minutes, the ratio of products to reactant concentration is 1:3. Thus, the ratio of remaining reactant to initial reactant is: \[ \frac{[A]}{[A]_0} = \frac{1}{4} \] Now applying the first-order rate equation: \[ \ln \left( \frac{[A]_0}{[A]} \right) = \ln(4) = kt \] Since \( \ln(4) = 1.386 \), we get: \[ 1.386 = k \cdot 24 \quad \Rightarrow \quad k = \frac{1.386}{24} = 0.05775 \, \text{min}^{-1} \] The half-life of a first-order reaction is given by: \[ t_{1/2} = \frac{0.693}{k} \] Substitute the value of \( k \): \[ t_{1/2} = \frac{0.693}{0.05775} = 12.0 \, \text{minutes} \] Thus, the half-life of the reaction is 157.8 minutes.