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?
Show Hint
- 60 min
- 80 min
- 20 min
- 40 min
The Correct Option is D
Solution and Explanation
\( \text{Given: } [A]_0 = 8[B]_0 \)
\( t_{1/2(A)} = 10 \text{ min} \)
\( t_{1/2(B)} = 40 \text{ min} \)
\( 1^{st} \text{ order kinetics} \)
\( t = ? \) \( [A] = [B] \) \( -k_A \times t = \ln \frac{[A]}{[A]_0} \)
\( [A] = [A]_0 e^{-k_A t} \) \( [B] = [B]_0 e^{-k_B t} \) \( [A] = [B] \)
\( [A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \)
\( 8[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \)
\( 8 = e^{(k_A - k_B)t} \) \( \ln 8 = (k_A - k_B)t \) \( t = \frac{\ln 8}{k_A - k_B} \)
\( t = \frac{\ln 8}{\frac{\ln 2}{10} - \frac{\ln 2}{40}} \)
\( t = \frac{\ln 2^3}{\frac{\ln 2}{10} - \frac{\ln 2}{40}} \)
\( t = \frac{3 \ln 2}{\ln 2 \left( \frac{1}{10} - \frac{1}{40} \right)} \)
\( t = \frac{3}{\frac{4-1}{40}} = \frac{3}{\frac{3}{40}} \)
\( t = 40 \text{ min} \)
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