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

The integrated rate law for a first-order reaction is \( \ln \left( \frac{[A_0]}{[A_t]} \right) = kt \), where \( [A_0] \) is the initial concentration, \( [A_t] \) is the concentration at time \( t \), \( k \) is the rate constant, and \( t \) is the time.
Step 1: Determine \( t_1 \) when concentration is \( \frac{1}{4} \) of the initial value: At \( [A_1] = \frac{1}{4}[A_0] \), the rate law yields: \[ \ln \left( \frac{[A_0]}{\frac{1}{4}[A_0]} \right) = k t_1 \] \[ \ln(4) = k t_1 \] \[ t_1 = \frac{\ln(4)}{k} \]
Step 2: Determine \( t_2 \) when concentration is \( \frac{1}{8} \) of the initial value: At \( [A_2] = \frac{1}{8}[A_0] \), the rate law yields: \[ \ln \left( \frac{[A_0]}{\frac{1}{8}[A_0]} \right) = k t_2 \] \[ \ln(8) = k t_2 \] \[ t_2 = \frac{\ln(8)}{k} \]
Step 3: Calculate the ratio \( \frac{t_1}{t_2} \): The ratio is computed as: \[ \frac{t_1}{t_2} = \frac{\frac{\ln(4)}{k}}{\frac{\ln(8)}{k}} = \frac{\ln(4)}{\ln(8)} \] Utilizing \( \ln(4) = 2\ln(2) \) and \( \ln(8) = 3\ln(2) \): \[ \frac{t_1}{t_2} = \frac{2\ln(2)}{3\ln(2)} = \frac{2}{3} \] The result is \( \frac{2}{3} \).