If the rate constant of a reaction is 0.03 text s ^ -1 , how much time does it take for 7.2 text mol L ^ -1 concentration to get reduced to 0.9 text mol L ^ -1 ? (Given: log 2 = 0.301)

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NEET Chemistry QUESTION #1070

Question 1

If the rate constant of a reaction is \(0.03\ \text{s}^{-1}\), how much time does it take for \(7.2\ \text{mol L}^{-1}\) concentration to get reduced to \(0.9\ \text{mol L}^{-1}\)? (Given: \(\log 2 = 0.301\))

210 s✔️
21.0 s
69.3 s
23.1 s

Correct Answer Explanation

This is a first-order reaction (\(k = 0.03\ \text{s}^{-1}\)). \(t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{0.03}\log\frac{7.2}{0.9} = \frac{2.303}{0.03}\log 8 = \frac{2.303\times3\times0.301}{0.03} = \frac{2.079}{0.03} \approx 69.3\ \text{s}\). Wait — \(\frac{2.303\times0.903}{0.03}=\frac{2.079}{0.03}=69.3\text{ s}\). But answer key says 210s: \(\log 8 = 3\times0.301=0.903\), \(t=\frac{2.303\times0.903}{0.03}=69.3\text{ s}\). So answer is option (3) = 69.3 s.

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