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: Understanding the Concept:
For a first order reaction the concentration falls exponentially, and the rate constant follows from the integrated rate law.

Step 2: Key Formula or Approach:
\[ k = \frac{2.303}{t} \log_{10}\frac{[A]_0}{[A]} \]

Step 3: Detailed Explanation:
20 % remains, so \([A] = 0.2[A]_0\) and \(\dfrac{[A]_0}{[A]} = 5\). Time \(t = 10\) min.
\[ k = \frac{2.303}{10} \times \log_{10} 5 = 0.2303 \times 0.6989 \]
\[ k = 0.1609\ \text{min}^{-1} \]

Step 4: Why the other options are wrong.
1.609, 6.989 and 16.09 are ten or more times too large. They come from dropping the division by 10 or a decimal slip. A first order reaction that loses 80 % in 10 minutes cannot have \(k\) above 1 per minute.

Final Answer:
The rate constant is 0.1609 \(\text{min}^{-1}\), option (D). \[ \boxed{0.1609\ \text{min}^{-1}} \]