Question

In first order reaction 20 millimole of reactant is reduced to 10 millimole in \(1\cdot 151\) minute. Find rate constant.

• A. \(0\cdot 6023\text{ minute}^{-1}\)
• B. \(6\cdot 120\text{ minute}^{-1}\)
• C. \(0\cdot 3010\text{ minute}^{-1}\)
• D. \(2\cdot 010\text{ minute}^{-1}\)

Hint:

Time taken for 50 percent change is the half-life, and k = 0.693 / t half.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Reactant falls from 20 to 10 millimole, which is half of the initial amount. So 1.151 minute is the half-life of this first order reaction.

Step 2: Key Formula or Approach:
\[ k = \frac{0.693}{t_{1/2}} = \frac{2.303\log 2}{t_{1/2}} \]

Step 3: Calculation:
\[ k = \frac{0.693}{1.151} = 0.602\text{ min}^{-1} \]
Using \(2.303\times0.3010 = 0.6932\), we get \(\frac{0.6932}{1.151} = 0.6023\text{ min}^{-1}\), option (A).

Final Answer:
The rate constant is 0.6023 per minute, option (A). \[ \boxed{0.6023\text{ min}^{-1}} \]