Question

The time taken for 60% completion of a first order reaction is 13.22 min. What is its half-life ($t_{1/2}$) in min? ($\log(2.5) = 0.398$)

• A. 11
• B. 9
• C. 8
• D. 10

Hint:

Notice that $2.303 \times \log(2) = 0.693$.
For first order reactions, you can relate any two times $t_1$ and $t_2$ directly:
$\frac{t_1}{t_2} = \frac{\log(a/(a-x_1))}{\log(a/(a-x_2))}$.
This bypasses calculating the rate constant $k$ explicitly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question is from chemical kinetics.
We are given the time required for 60% completion of a first-order chemical reaction and must find its half-life ($t_{1/2}$).

Step 2: Key Formula or Approach:
For a first-order reaction, the rate constant $k$ is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{a}{a - x} \right) \] where $a$ is the initial concentration and $x$ is the amount reacted at time $t$.
The half-life of a first-order reaction is related to $k$ by:
\[ t_{1/2} = \frac{0.693}{k} \]

Step 3: Detailed Explanation:

• Let us write down the given parameters:
Time ($t$) = 13.22 min.
Percentage of reaction completed = 60%.
Therefore, if $a = 100$, then $x = 60$, and the remaining reactant concentration is $a - x = 40$.

• Substitute these values into the first-order rate equation:
\[ k = \frac{2.303}{13.22} \log \left( \frac{100}{40} \right) \] \[ k = \frac{2.303}{13.22} \log (2.5) \]

• Using the given value $\log(2.5) = 0.398$:
\[ k = \frac{2.303 \times 0.398}{13.22} \] \[ k \approx \frac{0.9166}{13.22} \approx 0.06933\text{ min}^{-1} \]

• Now, we calculate the half-life ($t_{1/2}$) of the reaction:
\[ t_{1/2} = \frac{0.693}{k} \] \[ t_{1/2} = \frac{0.693}{0.06933} \approx 10\text{ min} \]

Step 4: Final Answer:
The half-life of the first-order reaction is 10 min.