Question

The time required for \(90\%\) completion of a certain first order reaction is \(1\) hour. Calculate the time required for \(99.9\%\) completion of the same reaction.

• A. \(2\) hours.
• B. \(1\) hour.
• C. \(3\) hours.
• D. \(0.5\) hour.

Hint:

t is proportional to log(A0/A). Compare log 10 and log 1000.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a first order reaction, \(k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\). If \(x\%\) has reacted, \([A] = (100-x)\%\) of \([A]_0\).

Step 2: For 90% completion:
\([A]_0/[A] = 100/10 = 10\), so
\[ k = \frac{2.303}{1\ \text{h}}\log 10 = 2.303\ \text{h}^{-1} \]

Step 3: For 99.9% completion:
\([A]_0/[A] = 100/0.1 = 1000\), so \(\log 1000 = 3\).
\[ t = \frac{2.303}{k}\times 3 = \frac{2.303\times 3}{2.303} = 3\ \text{h} \]

Step 4: Select:
The answer is 3 hours, option (C). The time is not simply proportional to the percentage.

Final Answer:
Time for 99.9 percent is three times the time for 90 percent, so 3 hours. \[ \boxed{3\ \text{hours}} \]