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Step-by-Step Solution
Step 1: Identify Known Information
• The half-life of substance A, t_{1/2,A} = 100\,\text{s} .
• The half-life of substance B, t_{1/2,B} = 50\,\text{s} .
• Initially, equal moles (and hence equal concentrations) of A and B are present.
• We want the time t when the concentration of A is four times the concentration of B.
Step 2: Write the Rate Constants
For a first-order reaction, the rate constant k is related to half-life by
k = \frac{\ln 2}{t_{1/2}}
Therefore,
k_{A} = \frac{\ln 2}{100}, \quad k_{B} = \frac{\ln 2}{50}.
Step 3: Write the Concentration Expressions
For first-order decay, the concentration at time t is given by
A_t = A_0 \, e^{-k_A \, t}, \quad B_t = B_0 \, e^{-k_B \, t}.
Since A_0 = B_0 (equal initial amounts), let that common initial concentration be C_0 .
Step 4: Apply the Given Condition (Concentration Ratio)
We want A_t = 4\,B_t . Substituting the expressions for A_t and B_t ,
C_0 \, e^{-k_A \, t} = 4 \, C_0 \, e^{-k_B \, t} \,.
Simplifying (since C_0 \neq 0 ),
e^{-k_A \, t} = 4 \, e^{-k_B \, t}.
Step 5: Solve for Time t
Divide both sides by e^{-k_B \, t} :
e^{-k_A \, t + k_B \, t} = 4 \,.
Notice that -k_A + k_B = -\frac{\ln 2}{100} + \frac{\ln 2}{50} = \frac{\ln 2}{100} . Thus, we get:
e^{\frac{\ln 2}{100} \, t} = 4 \,.
Taking the natural logarithm of both sides:
\frac{\ln 2}{100} \, t = \ln 4 = 2 \ln 2.
Hence,
t = \frac{2 \ln 2 \times 100}{\ln 2} = 200\,\text{s}.
Final Answer
The required time is 200 seconds.
