Question

A first order reaction take \(30\) minutes for \(75\%\) decomposition, calculate its rate constant ?

• A. \(0.0238 \text{minute}^{-1}\)
• B. \(0.0463 \text{minute}^{-1}\)
• C. \(0.0715 \text{minute}^{-1}\)
• D. \(0.0957 \text{minute}^{-1}\)

Hint:

Use k = (2.303/t) log (a/(a - x)); 75% decomposition leaves 25 percent.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a first-order reaction, \(k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\).

Step 2: Set Up the Data:
75% has decomposed, so 25% remains. If \([A]_0 = 100\), then \([A] = 25\) and the ratio is 4.
\(t = 30\) min.

Step 3: Calculation:
\[ k = \frac{2.303}{30}\log 4 = \frac{2.303\times 0.6021}{30} = \frac{1.3866}{30} = 0.0462\ \text{min}^{-1} \approx 0.0463\ \text{min}^{-1} \]

Step 4: Check the Other Options:
0.0238 and 0.0715 and 0.0957 min\(^{-1}\) do not satisfy \(kt = \ln 4 = 1.386\) for \(t = 30\) min. For 0.0463, \(kt = 1.389\), which matches. So (B) is correct.

Final Answer:
The rate constant is \(0.0463\ \text{min}^{-1}\), option (B). \[ \boxed{\text{(B) } 0.0463\ \text{min}^{-1}} \]