Question

At 30ºC, the half life for the decomposition of \(AB_2\) is 200 s and is independent of the initial concentration of \(AB_2\). The time required for 80% of the \(AB_2\) to decompose is(Given : \(log\; 2 = 0.30,\; log \;3 = 0.48\))

• A. 200 s
• B. 323 s
• C. 467 s
• D. 532 s

Answer 1

The Correct Option is C

The given problem involves the decomposition of a compound \(AB_2\) and requires calculating the time needed for 80% decomposition. The problem provides the half-life of the decomposition reaction as being independent of the initial concentration, indicating a first-order reaction. Let's solve step-by-step:

For a first-order reaction, the half-life (\(t_{1/2}\)) is given by the formula:

\(t_{1/2} = \frac{0.693}{k}\)

Where \(k\) is the rate constant.

Substitute the given half-life into the formula:

\(200 = \frac{0.693}{k}\)

Solving for \(k\):

\(k = \frac{0.693}{200} = 0.003465\; \text{s}^{-1}\)

For a first-order reaction, the time taken for a certain fraction of the reaction to occur is given by:

\(t = \frac{2.303}{k} \log \left(\frac{[A_0]}{[A]}\right)\)

Where \([A_0]\) is the initial concentration and \([A]\) is the concentration at time \(t\).

For 80% decomposition, 20% of the original compound \(AB_2\) remains, so \(\frac{[A_0]}{[A]} = \frac{100}{20} = 5\).

Substitute the values into the equation:

\(t = \frac{2.303}{0.003465} \log (5)\)

Given: \(\log 2 = 0.30\) and \(\log 3 = 0.48\).

Calculate \(\log 5\):

\(\log 5 = \log (10/2) = \log 10 - \log 2 = 1 - 0.30 = 0.70\)

Now, substitute \(\log 5\) and calculate \(t\):

\(t = \frac{2.303}{0.003465} \times 0.70 \approx 467\; \text{s}\)

Thus, the correct time required for 80% of \(AB_2\) to decompose is 467 seconds.