Question

The rate constant for a First order reaction at 560 K is $1.5 \times 10^{-6}$ per secon(D) If the reaction is allowed to take place for 20 hours, what percentage of the initial concentration would have converted to products?

• A. 11.14 %
• B. 10.23 %
• C. 12.46 %
• D. 21.2 %

Hint:

For first-order reactions, the time taken to reach a certain percentage of completion can be calculated using the equation \( \ln \left( \frac{[A_0]}{[A]} \right) = kt \). The percentage reacted is directly related to the change in the concentration of the reactant.

Answer 1

The Correct Option is B

For a first-order reaction, the relationship between concentration and time is given by the equation: \[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \] where: - \([A_0]\) is the initial concentration, - \([A]\) is the concentration at time \(t\), - \(k\) is the rate constant, and - \(t\) is the time. We are given: - \(k = 1.5 \times 10^{-6} \ \text{s}^{-1}\), - \(t = 20 \ \text{hours} = 20 \times 3600 \ \text{seconds} = 72,000 \ \text{seconds}\). We need to find the percentage of the initial concentration that has reacte(D) This is related to the change in concentration: \[ \text{Percentage reacted} = \frac{[A_0] - [A]}{[A_0]} \times 100 \] Using the first-order equation, we calculate the remaining concentration at time \(t\): \[ \ln \left( \frac{[A_0]}{[A]} \right) = k \times t \] Substituting the values: \[ \ln \left( \frac{[A_0]}{[A]} \right) = (1.5 \times 10^{-6}) \times 72,000 \] \[ \ln \left( \frac{[A_0]}{[A]} \right) = 0.108 \] Now solving for \(\frac{[A_0]}{[A]}\): \[ \frac{[A_0]}{[A]} = e^{0.108} \approx 1.114 \] Thus, the remaining concentration \([A]\) is: \[ [A] = \frac{[A_0]}{1.114} \] Now, the percentage reacted is: \[ \text{Percentage reacted} = \left(1 - \frac{1}{1.114} \right) \times 100 \approx 10.23 % \] Therefore, the correct answer is 10.23 %.