Question

For a first order reaction, time required for 99% completion is \(x\) times the time required for 90% completion. Find \(x\).

Hint:

For first order reactions, \(t_{99\%} = 2 \times t_{90\%}\) because \(\log 100 = 2 \log 10\). In general, \(t_{99.9\%} = 3 \times t_{90\%}\), etc.

Answer 1

Correct Answer: 2

Concept: For first order reaction: \[ t = \frac{2.303}{k} \log \frac{a}{a-x} \] where \(a\) = initial concentration, \(x\) = amount reacted.

Step 1:For 90% completion: \[ \frac{x}{a} = 0.90 \Rightarrow \frac{a}{a-x} = \frac{100}{10} = 10 \] \[ t_{90} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} \times 1 = \frac{2.303}{k} \]

Step 2:For 99% completion: \[ \frac{x}{a} = 0.99 \Rightarrow \frac{a}{a-x} = \frac{100}{1} = 100 \] \[ t_{99} = \frac{2.303}{k} \log 100 = \frac{2.303}{k} \times 2 = \frac{2 \times 2.303}{k} \]

Step 3:Ratio: \[ x = \frac{t_{99}}{t_{90}} = \frac{\frac{2 \times 2.303}{k}}{\frac{2.303}{k}} = 2 \] \[ \Rightarrow x = 2 \]