Question

If compound A reacts with B following first-order kinetics with rate constant \(2.011 \times 10^{-3} \, \text{s}^{-1}\), the time taken by A (in seconds) to reduce from 7 g to 2 g will be ------- (Nearest Integer).
Given: \[\log 5 = 0.698, \, \log 7 = 0.845, \, \log 2 = 0.301.\]

Answer 1

Correct Answer: 623

Step 1: Reaction and First-Order Kinetics Formula
The reaction is:
\[\text{A} + \text{B} \rightarrow \text{P}.\]
At \(t = 0\), the concentration of A is \(7 \, \text{g}\). At \(t = t\), the concentration of A reduces to \(2 \, \text{g}\). For first-order reactions:
\[t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}.\]
Step 2: Substitute the Values
Substitute \(k = 2.011 \times 10^{-3} \, \text{s}^{-1}\), \([A]_0 = 7\), and \([A]_t = 2\):
\[t = \frac{2.303}{2.011 \times 10^{-3}} \log \frac{7}{2}.\]
\[\log \frac{7}{2} = \log 7 - \log 2 = 0.845 - 0.301 = 0.544.\]
Step 3: Calculate the Time
Substitute the values:
\[t = \frac{2.303}{2.011 \times 10^{-3}} \cdot 0.544.\]
\[t = \frac{2.303 \times 0.544}{2.011 \times 10^{-3}} = \frac{1.252832}{2.011 \times 10^{-3}}.\]
\[t = 622.989 \, \text{seconds} \approx 623 \, \text{seconds}.\]
Conclusion: The time taken by A to reduce from 7 g to 2 g is \(\mathbf{623 \, \text{seconds}}\).