Question

The half-life of a first order reaction is 6.0 hours. How long does it take for 60% of the reactant to disappear?

• A. 7.93 hours
• B. 12.0 hours
• C. 15.6 hours
• D. 10.43 hours

Hint:

Kinetics Tip: Always distinguish between "amount reacted" and "amount remaining". The Nernst and Rate equations always use the "amount remaining" in the denominator.

Answer 1

The Correct Option is A

Concept: Chemistry (Chemical Kinetics) - First Order Reaction Dynamics.

Step 1: Calculate the rate constant ($k$). For a first-order reaction, the relationship between half-life ($t_{1/2}$) and the rate constant is $k = \frac{0.693}{t_{1/2}}$. Given $t_{1/2} = 6.0$ hours: $k = \frac{0.693}{6.0} = 0.1155 \text{ hr}^{-1}$.

Step 2: Define the initial and final concentrations. Let the initial concentration $[A]_0$ be 100. The question states that 60% of the reactant "disappears". This means 60% has reacted. The remaining concentration $[A]_t = 100 - 60 = 40$.

Step 3: Set up the first-order integrated rate equation. The formula to find the time ($t$) is: $t = \frac{2.303}{k} \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$.

Step 4: Substitute the values into the equation. $t = \frac{2.303}{0.1155} \log_{10} \left( \frac{100}{40} \right)$. Simplify the log term: $\log_{10}(2.5)$. Using standard log tables, $\log_{10} 2.5 \approx 0.3979$.

Step 5: Calculate the final time. $t = \frac{2.303 \times 0.3979}{0.1155} \approx \frac{0.9164}{0.1155} \approx 7.934$ hours. This value matches option A. $$ \therefore \text{It takes 7.93 hours for 60\% of the reactant to disappear.} $$