Question

What is rate constant of a first order reaction if 0.08 mole of reactant reduces to 0.02 mole in 23.03 minute?

• A. $0.2303\ \text{min}^{-1}$
• B. $1.6021\ \text{min}^{-1}$
• C. $0.4031\ \text{min}^{-1}$
• D. $0.06021\ \text{min}^{-1}$

Hint:

Notice that the initial amount ($0.08\ \text{mol}$) gets cut in half twice to become $0.02\ \text{mol}$ ($0.08 \rightarrow 0.04 \rightarrow 0.02$), meaning exactly two half-lives ($2 \times t_{1/2}$) have passed. $$2 \times t_{1/2} = 23.03 \implies t_{1/2} = 11.515\ \text{min}$$ Then, use $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{11.515} \approx 0.0602\ \text{min}^{-1}$ as an alternative shortcut.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the initial and final quantities of a reactant along with the time elapsed for a first-order chemical reaction, and we need to calculate its specific rate constant ($k$).

Step 2: Key Formula or Approach:
The integrated rate equation for a first-order reaction is expressed as: $$k = \frac{2.303}{t} \log_{10}\left(\frac{[A]_0}{[A]_t}\right)$$ where $[A]_0$ is the initial concentration or amount, and $[A]_t$ is the final concentration or amount remaining at time $t$.

Step 3: Detailed Explanation:
Given values: Initial amount, $[A]_0 = 0.08\ \text{mol}$ Remaining amount, $[A]_t = 0.02\ \text{mol}$ Time elapsed, $t = 23.03\ \text{min}$ Substitute these parameters into the first-order rate formula: $$k = \frac{2.303}{23.03} \log_{10}\left(\frac{0.08}{0.02}\right)$$ Simplify the leading fraction and the logarithmic argument: $$k = 0.1 \times \log_{10}(4)$$ Since $\log_{10}(4) = 2 \times \log_{10}(2) \approx 2 \times 0.3010 = 0.6021$: $$k = 0.1 \times 0.6021 = 0.06021\ \text{min}^{-1}$$

Step 4: Final Answer:
The calculated value for the rate constant is $0.06021\ \text{min}^{-1}$, matching option (D).