Question

In a first order reaction 20 millimole of reactant is lowered to 10 millimole in \(0.3010\) minute. Find rate constant of the reaction?

• A. \(3.010 \text{minute}^{-1}\)
• B. \(0.602 \text{minute}^{-1}\)
• C. \(2.303 \text{minute}^{-1}\)
• D. \(0.301 \text{minute}^{-1}\)

Hint:

Reactant halves in 0.3010 min, so this is the half life.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a first order reaction the time for the concentration to fall to half is a constant, called the half life: \(t_{1/2} = \frac{0.693}{k}\).

Step 2: Key Formula or Approach:
The reactant goes from \(20\) to \(10\) millimole, which is exactly half, so \(t_{1/2} = 0.3010\) min.

Step 3: Detailed Explanation:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{0.3010} = 2.303\ \text{min}^{-1} \]
Check with the integrated rate law: \(k = \frac{2.303}{t}\log\frac{20}{10} = \frac{2.303 \times 0.3010}{0.3010} = 2.303\) min\(^{-1}\).
Option D, \(0.301\), is just the value of \(\log 2\) and not a rate constant. Options A and B come from slips in the arithmetic.

Final Answer:
The rate constant is \(2.303\) min\(^{-1}\), option (C). \[ \boxed{2.303\ \text{min}^{-1}} \]