Question

In a first order reaction, the concentration of a reactant decreases from $20\ \text{mmol}$ to $10\ \text{mmol}$ in $1.151\ \text{min}$. What is the rate constant?

• A. $1.15\ \text{min}^{-1}$
• B. $3.0\ \text{min}^{-1}$
• C. $5.50\ \text{min}^{-1}$
• D. $0.60\ \text{min}^{-1}$

Hint:

Recognizing that dropping from 20 to 10 signifies exactly one half-life saves calculation time. Since $t_{1/2} = 1.151\ \text{min}$, simply evaluating $\frac{0.693}{1.151}$ reveals it must be slightly less than $0.7$, leading directly to option (D).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to find the specific chemical rate constant ($k$) for a first-order kinetics process where the reactant concentration drops to exactly half of its initial amount within a span of $1.151\ \text{min}$.

Step 2: Key Formula or Approach:
The integrated rate equation tracking a first-order chemical process is given by:
$$ k = \frac{2.303}{t} \log_{10} \left( \frac{[\text{A}]_0}{[\text{A}]_t} \right) $$ Alternatively, because the concentration drops from $20\ \text{mmol}$ to $10\ \text{mmol}$, this elapsed timeframe corresponds to the reaction's half-life ($t_{1/2}$). The half-life formula is:
$$ k = \frac{0.693}{t_{1/2}} $$

Step 3: Detailed Explanation:
Let's substitute the given values into the primary integrated equation:
Initial concentration $[\text{A}]_0 = 20\ \text{mmol}$
Final concentration $[\text{A}]_t = 10\ \text{mmol}$
Time interval $t = 1.151\ \text{min}$
$$ k = \frac{2.303}{1.151} \log_{10} \left( \frac{20}{10} \right) $$ Notice that $\frac{2.303}{1.151}$ reduces cleanly to exactly $2$:
$$ k = 2 \times \log_{10}(2) $$ Since $\log_{10}(2) \approx 0.3010$:
$$ k = 2 \times 0.3010 = 0.602\ \text{min}^{-1} \approx 0.60\ \text{min}^{-1} $$

Step 4: Final Answer: The chemical rate constant matches option (D).