Question

For a first order reaction at \(27^\circ C\), ratio of time required for \(75%\) completion to \(25%\) completion of reaction is

• A. \(3.0\)
• B. \(2.303\)
• C. \(4.8\)
• D. \(0.477\)

Hint:

For first order reactions, time depends on \(\log\left(\frac{a}{a-x}\right)\). Higher completion needs disproportionately larger time.

Answer 1

The Correct Option is C

Step 1: Use first order time formula.
For first order reaction:
\[ t = \frac{2.303}{k}\log\left(\frac{a}{a-x}\right) \]
Step 2: Time for 25% completion.
\[ x = 0.25a \Rightarrow a-x = 0.75a \]
\[ t_{25} = \frac{2.303}{k}\log\left(\frac{a}{0.75a}\right) = \frac{2.303}{k}\log\left(\frac{4}{3}\right) \]
Step 3: Time for 75% completion.
\[ x = 0.75a \Rightarrow a-x = 0.25a \]
\[ t_{75} = \frac{2.303}{k}\log\left(\frac{a}{0.25a}\right) = \frac{2.303}{k}\log(4) \]
Step 4: Ratio of times.
\[ \frac{t_{75}}{t_{25}} = \frac{\log(4)}{\log(4/3)} \]
\[ \log(4)=0.6021,\quad \log(4/3)=0.1249 \]
\[ \frac{t_{75}}{t_{25}} \approx \frac{0.6021}{0.1249} \approx 4.82 \]
So:
\[ \boxed{4.8} \]
Final Answer:
\[ \boxed{4.8} \]