Question

In a reaction A + B → C, initial concentrations of A and B are related as $ [\text{A}]_0 = 8[\text{B}]_0 $. The half lives of A and B are 10 min and 40 min, respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

• A. 60 min
• B. 80 min
• C. 20 min
• D. 40 min

Hint:

Use the first-order rate law and the given half-lives to find the time when the concentrations of A and B are equal.

Answer 1

The Correct Option is D

\( \text{Given: } [A]_0 = 8[B]_0 \)

\( t_{1/2(A)} = 10 \text{ min} \) 

\( t_{1/2(B)} = 40 \text{ min} \)

 \( 1^{st} \text{ order kinetics} \) 

\( t = ? \) \( [A] = [B] \) \( -k_A \times t = \ln \frac{[A]}{[A]_0} \) 

\( [A] = [A]_0 e^{-k_A t} \) \( [B] = [B]_0 e^{-k_B t} \) \( [A] = [B] \) 

\( [A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \) 

\( 8[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \) 

\( 8 = e^{(k_A - k_B)t} \) \( \ln 8 = (k_A - k_B)t \) \( t = \frac{\ln 8}{k_A - k_B} \) 

\( t = \frac{\ln 8}{\frac{\ln 2}{10} - \frac{\ln 2}{40}} \) 

\( t = \frac{\ln 2^3}{\frac{\ln 2}{10} - \frac{\ln 2}{40}} \) 

\( t = \frac{3 \ln 2}{\ln 2 \left( \frac{1}{10} - \frac{1}{40} \right)} \) 

\( t = \frac{3}{\frac{4-1}{40}} = \frac{3}{\frac{3}{40}} \) 

\( t = 40 \text{ min} \)