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Step-by-Step Solution
Step 1: Understand the Problem
We have two compounds, A and B, both decomposing according to first-order kinetics. The half-life of A is 300 s, and the half-life of B is 180 s. Initially, their concentrations are equal. We need to find the time at which the concentration of A becomes four times that of B.
Step 2: Write the Rate Constants
The rate constant $k$ for a first-order reaction and its half-life $t_{1/2}$ follow the relationship:
$k = \frac{\ln 2}{t_{1/2}}.$
Hence:
$k_1 = \frac{\ln 2}{300} \quad \text{for compound A}, \quad
k_2 = \frac{\ln 2}{180} \quad \text{for compound B}.
Step 3: Express Concentrations Using First-Order Kinetics
For a first-order reaction, the concentration at time $t$ is given by:
$[A]_t = [A]_0 \, e^{-k_1 t}, \quad [B]_t = [B]_0 \, e^{-k_2 t}.
Since initially the concentrations are equal, let $[A]_0 = [B]_0.$
Step 4: Set Up the Condition for the Desired Concentration Ratio
We want $[A]_t$ to be four times $[B]_t$:
$[A]_t = 4\,[B]_t.
Substitute the expressions for $[A]_t$ and $[B]_t$:
$[A]_0 \, e^{-k_1 t} = 4 \,[B]_0 \, e^{-k_2 t}.
Because $[A]_0 = [B]_0,$ we can divide both sides by $[A]_0$:
$e^{-k_1 t} = 4\,e^{-k_2 t}.
Rearrange to:
$e^{(k_2 - k_1)t} = 4.
Step 5: Solve for t
1. Take the natural logarithm on both sides:
$(k_2 - k_1)\,t = \ln(4) = 2 \ln(2).
2. Substitute $k_1 = \frac{\ln 2}{300}$ and $k_2 = \frac{\ln 2}{180}$:
$\left(\frac{\ln 2}{180} - \frac{\ln 2}{300}\right) t = 2 \ln 2.
3. Factor out $\ln 2$:
$(\ln 2)\left(\frac{1}{180} - \frac{1}{300}\right) t = 2 (\ln 2).
4. Cancel $\ln 2$ on both sides:
$\left(\frac{1}{180} - \frac{1}{300}\right) t = 2.
5. Find the common denominator and simplify:
$\frac{1}{180} = \frac{5.555...}{1000},\; \frac{1}{300} = \frac{3.333...}{1000}, \;
\text{so} \; \left(\frac{1}{180} - \frac{1}{300}\right) = \frac{1}{180} - \frac{1}{300}
= \frac{5 - 3}{900} = \frac{2}{900} = \frac{1}{450}.$
Thus,
$\frac{1}{450}\,t = 2 \quad \Longrightarrow \quad t = 2 \times 450 = 900 \text{ s}.
Answer
The required time is 900 seconds.
