Half-Life & Concentration Ratios
Compare the decay of two different substances to find a specific concentration ratio.
Question:
A flask is filled with equal moles of $A$ and $B$. The half lives of $A$ and $B$ are $100 \text{ s}$ and $50 \text{ s}$ respectively and are independent of the initial concentration.
The time required for the concentration of $A$ to be four times that of $B$ is ________ $\text{s}$.
[Given: $\ln 2 = 0.693$]
Detailed Solution
Step 1: Set up the equations
Let the initial concentration of both $A$ and $B$ be $N_0$. Let the required time be $t$.
- Concentration of $A$ at time $t$: $\quad N_A = \frac{N_0}{2^{t/100}}$
- Concentration of $B$ at time $t$: $\quad N_B = \frac{N_0}{2^{t/50}}$
Step 2: Apply the given condition
We are given that at time $t$, the concentration of $A$ is four times that of $B$: $N_A = 4 \times N_B$.
Cancel $N_0$ from both sides and rearrange to group the exponent terms:
$$ 2^{\left(\frac{t}{50} - \frac{t}{100}\right)} = 2^2 $$
Since the bases ($2$) are the same, we can equate the exponents:
$$ \frac{2t - t}{100} = 2 $$
$$ \frac{t}{100} = 2 \implies t = 200 \text{ s} $$
The Mental Math Shortcut
Since $A$'s half-life is $100 \text{ s}$ and $B$'s is $50 \text{ s}$, $B$ is decaying twice as fast. Let's just track their amounts in $100 \text{ s}$ intervals! Let initial amount = $1$.
| Time ($t$) | Amount of $A$ ($t_{1/2} = 100$s) | Amount of $B$ ($t_{1/2} = 50$s) | Ratio ($A/B$) |
|---|---|---|---|
| $0 \text{ s}$ | 1 | 1 | 1 |
| $100 \text{ s}$ | $0.5$ (1 HL) | $0.25$ (2 HLs) | 2 |
| $200 \text{ s}$ | $0.25$ (2 HLs) | $0.0625$ (4 HLs) | 4 |
At 200 seconds, $0.25$ is exactly $4$ times $0.0625$. No complex algebra needed!
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