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Step-by-Step Solution
Step 1: Identify the given data
• Half-life of A, T_{1/2}(A) = 54 \text{ min}
• Half-life of B, T_{1/2}(B) = 18 \text{ min}
• Initial concentrations: [A_{0}] = [B_{0}] = a
• We need the time t such that [A] = 16 \, [B] .
Step 2: Write the concentration expressions for first-order kinetics
For a first-order reaction, the concentration after time t is:
[A] = [A_0] e^{-k_A t} and [B] = [B_0] e^{-k_B t} .
Step 3: Relate the half-life to the rate constant
For a first-order reaction, k = \frac{\ln 2}{T_{1/2}} . Hence:
k_A = \frac{\ln 2}{54} \quad \text{and} \quad k_B = \frac{\ln 2}{18}.
Step 4: Apply the condition [A] = 16 \, [B]
Using the expressions:
[A_0] e^{-k_A t} = 16 \, [B_0] e^{-k_B t}.
Since [A_0] = [B_0] = a , we get:
a \, e^{-k_A t} = 16 \, a \, e^{-k_B t},
which simplifies to
e^{-\left(k_A - k_B\right) t} = 16.
Step 5: Solve for t
Taking the natural logarithm on both sides:
- (k_A - k_B)\,t = \ln(16).
Since \ln(16) = \ln(2^4) = 4 \ln(2) , we have:
(k_B - k_A) \, t = 4 \ln(2).
Substitute k_A = \frac{\ln 2}{54} and k_B = \frac{\ln 2}{18} :
k_B - k_A = \frac{\ln 2}{18} - \frac{\ln 2}{54} = \ln(2)\left(\frac{1}{18} - \frac{1}{54}\right) = \ln(2)\left(\frac{3 - 1}{54}\right) = \ln(2)\left(\frac{2}{54}\right) = \frac{\ln(2)}{27}.
So,
\frac{\ln(2)}{27} \, t = 4 \ln(2).
Therefore,
t = \frac{4 \ln(2)}{\frac{\ln(2)}{27}} = 4 \times 27 = 108 \text{ min}.
Final Answer
The required time is 108 minutes.
