Question:
For a reaction \(\text{A}⟶\) product, \(k = 2\times 10^{-2} \text{s}^{-1}\). If the initial concentration of A is \(1.0 \text{mol dm}^{-3}\) find the value of \(log\frac{1}{[A]_t}\) after 100 second ?
For a reaction \(\text{A}⟶\) product, \(k = 2\times 10^{-2} \text{s}^{-1}\). If the initial concentration of A is \(1.0 \text{mol dm}^{-3}\) find the value of \(log\frac{1}{[A]_t}\) after 100 second ?
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A reaction with unit of k in s^-1 is first order. Use log[A]0/[A] = kt/2.303.
Updated On: Oct 1, 2026
- \(0.135\)
- \(0.270\)
- \(0.430\)
- \(0.868\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
The unit of \(k\) is \(\text{s}^{-1}\), which tells us the reaction is first order. For a first order reaction the concentration falls exponentially with time.
Step 2: Key Formula or Approach:
\[ k = \frac{2.303}{t}\log\frac{[A]_0}{[A]_t} \;\Rightarrow\; \log\frac{[A]_0}{[A]_t} = \frac{kt}{2.303} \]
Since \([A]_0 = 1.0\), the left side equals \(\log\dfrac{1}{[A]_t}\).
Step 3: Detailed Explanation:
\[ \log\frac{1}{[A]_t} = \frac{2\times10^{-2}\times100}{2.303} = \frac{2}{2.303} = 0.868 \]
This means \([A]_t = 10^{-0.868} = 0.135 \text{ mol dm}^{-3}\). Option (A) 0.135 is that concentration and not the logarithm asked for. Option (B) 0.270 is double 0.135, and option (C) 0.430 comes from using \(kt = 1\), neither of which answers the question.
Final Answer:
The value of \(\log\frac{1}{[A]_t}\) is 0.868, option (D). \[ \boxed{0.868 \text{ (D)}} \]
The unit of \(k\) is \(\text{s}^{-1}\), which tells us the reaction is first order. For a first order reaction the concentration falls exponentially with time.
Step 2: Key Formula or Approach:
\[ k = \frac{2.303}{t}\log\frac{[A]_0}{[A]_t} \;\Rightarrow\; \log\frac{[A]_0}{[A]_t} = \frac{kt}{2.303} \]
Since \([A]_0 = 1.0\), the left side equals \(\log\dfrac{1}{[A]_t}\).
Step 3: Detailed Explanation:
\[ \log\frac{1}{[A]_t} = \frac{2\times10^{-2}\times100}{2.303} = \frac{2}{2.303} = 0.868 \]
This means \([A]_t = 10^{-0.868} = 0.135 \text{ mol dm}^{-3}\). Option (A) 0.135 is that concentration and not the logarithm asked for. Option (B) 0.270 is double 0.135, and option (C) 0.430 comes from using \(kt = 1\), neither of which answers the question.
Final Answer:
The value of \(\log\frac{1}{[A]_t}\) is 0.868, option (D). \[ \boxed{0.868 \text{ (D)}} \]
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