Question

Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction. ($k=0.02303\text{ minute}^{-1}$)

• A. 25%
• B. 75%
• C. 50%
• D. 12.5%

Hint:

Logic Tip: You can also use the integrated rate equation directly: $k = \frac{2.303}{t} \log\left(\frac{100}{100-x}\right)$. Substituting $k=0.02303$ and $t=60$ yields $0.02303 = \frac{2.303}{60} \log\left(\frac{100}{N_t}\right) \implies \log\left(\frac{100}{N_t}\right) = 0.6 \implies \frac{100}{N_t} \approx 4 \implies N_t = 25%$. But the half-life method is much faster!

Answer 1

The Correct Option is A

Concept:
For a first-order reaction, the half-life ($t_{1/2}$) is independent of the initial concentration and depends only on the rate constant ($k$) via the formula $t_{1/2} = \frac{0.693}{k}$. The amount of reactant remaining after $n$ half-lives can be easily calculated using the formula $\text{Remaining} = \frac{100%}{2^n}$.
Step 1: Calculate the half-life ($t_{1/2}$) of the reaction.
Given the rate constant $k = 0.02303\text{ min}^{-1}$. $$t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{0.02303}$$ $$t_{1/2} = \frac{0.693}{0.02303} \approx 30.09\text{ minutes}$$ For practical calculation purposes, $t_{1/2} = 30\text{ minutes}$.
Step 2: Determine the number of half-lives that have passed.
The total time given is $t = 60\text{ minutes}$. Number of half-lives, $n = \frac{\text{Total Time{t_{1/2$ $$n = \frac{60}{30} = 2$$ Exactly 2 half-lives have elapsed.
Step 3: Calculate the percentage of reactant remaining.
Initial concentration = 100% After 1st half-life (30 mins): $100% \div 2 = 50%$ remains. After 2nd half-life (60 mins): $50% \div 2 = 25%$ remains. Alternatively, using the formula: $$\text{Remaining Percentage} = \frac{100%}{2^n} = \frac{100%}{2^2} = \frac{100%}{4} = 25%$$