Question

At 30$^\circ$C, the half-life for the decomposition of a compound is 100 seconds and the value of the half-life is independent of the initial concentration of the reactantThe time required for 30\% of the reactant to be consumed is _ _ _ seconds. (rounded off to one decimal place)

Hint:

For first order reactions, use $\ln\left(\frac{[A]_0}{[A]}\right)=kt$ directly when percentage remaining is given

Answer 1

Correct Answer: 51.4

Step 1: Identify order of reaction.
Since half-life is independent of initial concentration, the reaction is first order

Step 2: Use half-life relation for first order reaction.
\[ t_{1/2} = \frac{0.693}{k} \]
\[ 100 = \frac{0.693}{k} \Rightarrow k = \frac{0.693}{100} = 0.00693 \, \text{s}^{-1} \]

Step 3: Determine fraction remaining.
30\% is consumed, so 70\% remains
\[ \frac{[A]}{[A]_0} = 0.70 \]

Step 4: Use first order rate law.
\[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \]
\[ \ln \left(\frac{1}{0.70}\right) = kt \]

Step 5: Solve for time.
\[ t = \frac{\ln(1/0.70)}{0.00693} \]
\[ t = \frac{0.3567}{0.00693} \]
\[ t \approx 51.4 \text{ s} \]

Step 6: Conclusion.
\[ \boxed{51.4 \text{ seconds}} \]