Question

In first order reaction, 20 % concentration remains after 10 min. What would be the rate constant of reaction if \(log_{10}(5) = 0.6989\)?

• A. \(1.609\) min\(^{-1}\)
• B. \(6.989\) min\(^{-1}\)
• C. \(16.09\) min\(^{-1}\)
• D. \(0.1609\) min\(^{-1}\)

Hint:

Use the first order integrated rate equation with [A]0/[A] = 5.

Answer 1

The Correct Option is D

Step 1: Use the natural log form.
$\ln\dfrac{[A]_0}{[A]} = kt$, so $k = \dfrac{\ln 5}{10}$.

Step 2: Convert.
$\ln 5 = 2.303 \times \log_{10} 5 = 2.303 \times 0.6989 = 1.6095$.

Step 3: Divide by time.
\[ k = \frac{1.6095}{10} = 0.1609\ \text{min}^{-1} \]

Step 4: Check.
$e^{-0.1609 \times 10} = e^{-1.609} = 0.2$, which matches 20 % remaining.

Final Answer:
Option (D) is correct. \[ \boxed{0.1609\ \text{min}^{-1}} \]