Question

For a first-order reaction, the ratio of time required is $ \frac{t_1}{t_2} $, if $ t_1 $ is the time consumed when reactant reaches $ \frac{1}{4} $ of initial concentration and $ t_2 $ is the time when it reaches $ \frac{1}{8} $ of initial concentration.

• A. \( \frac{2}{3} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{3}{2} \)
• D. \( \frac{4}{3} \)

Hint:

For first-order reactions, the ratio of times when the concentration changes to different fractions of the initial concentration can be found by using the relationship between the logarithms of those fractions.

Answer 1

The Correct Option is A

For a first-order reaction, the integrated rate law is given by: \[ \ln \left( \frac{[A_0]}{[A_t]} \right) = kt \] where: - \( [A_0] \) is the initial concentration, - \( [A_t] \) is the concentration of reactant at time \( t \), - \( k \) is the rate constant, - \( t \) is the time taken.
Step 1: Calculate \( t_1 \) when concentration reaches \( \frac{1}{4} \) of its initial value: For \( [A_1] = \frac{1}{4}[A_0] \), the equation becomes: \[ \ln \left( \frac{[A_0]}{\frac{1}{4}[A_0]} \right) = k t_1 \] \[ \ln(4) = k t_1 \] \[ t_1 = \frac{\ln(4)}{k} \]
Step 2: Calculate \( t_2 \) when concentration reaches \( \frac{1}{8} \) of its initial value: For \( [A_2] = \frac{1}{8}[A_0] \), the equation becomes: \[ \ln \left( \frac{[A_0]}{\frac{1}{8}[A_0]} \right) = k t_2 \] \[ \ln(8) = k t_2 \] \[ t_2 = \frac{\ln(8)}{k} \]
Step 3: Calculate the ratio \( \frac{t_1}{t_2} \): Now, we calculate the ratio \( \frac{t_1}{t_2} \): \[ \frac{t_1}{t_2} = \frac{\frac{\ln(4)}{k}}{\frac{\ln(8)}{k}} = \frac{\ln(4)}{\ln(8)} \] Since \( \ln(4) = 2\ln(2) \) and \( \ln(8) = 3\ln(2) \): \[ \frac{t_1}{t_2} = \frac{2\ln(2)}{3\ln(2)} = \frac{2}{3} \] Thus, the correct answer is \( \frac{2}{3} \).