Question:
At 298 K, for a first order reaction (A → P) the following graph is obtained. The rate constant (in s\(^{-1}\)) and initial concentration (in mol L\(^{-1}\)) of ‘A’ are respectively:
At 298 K, for a first order reaction (A → P) the following graph is obtained. The rate constant (in s\(^{-1}\)) and initial concentration (in mol L\(^{-1}\)) of ‘A’ are respectively:
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For first-order reactions, the slope of the \(\ln[A]\) vs. time graph gives the rate constant \( k \).
Updated On: May 16, 2025
- \( 2.303; 10^{-1} \)
- \( 10^{-2}; 2.303 \)
- \( 10^{-1}; 10^{-2} \)
- \( 10^{-2}; 10^{-1} \)
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The Correct Option is D
Solution and Explanation
Step 1: Identifying the Rate Constant.
From the integrated rate equation for a first-order reaction:
\[ \ln[A] = -kt + \ln[A]_0 \]
- The slope of the graph is given as \( -10^{-2} \), which corresponds to the rate constant: \[ k = 10^{-2} \text{ s}^{-1} \] Step 2: Identifying the Initial Concentration.
- The y-intercept of the graph corresponds to \( \ln[A]_0 \). Given \( \ln[A]_0 = -2.303 \), \[ [A]_0 = e^{-2.303} = 10^{-1} \text{ mol L}^{-1} \] Thus, the correct answer is \( k = 10^{-2} \text{ s}^{-1} \) and \( [A]_0 = 10^{-1} \text{ mol L}^{-1} \).
From the integrated rate equation for a first-order reaction:
\[ \ln[A] = -kt + \ln[A]_0 \]
- The slope of the graph is given as \( -10^{-2} \), which corresponds to the rate constant: \[ k = 10^{-2} \text{ s}^{-1} \] Step 2: Identifying the Initial Concentration.
- The y-intercept of the graph corresponds to \( \ln[A]_0 \). Given \( \ln[A]_0 = -2.303 \), \[ [A]_0 = e^{-2.303} = 10^{-1} \text{ mol L}^{-1} \] Thus, the correct answer is \( k = 10^{-2} \text{ s}^{-1} \) and \( [A]_0 = 10^{-1} \text{ mol L}^{-1} \).
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