Question:
For Arrhenius model of the rate constant \( k = k_0 \cdot e^{-E/RT} \), the graph drawn between \( \ln(k) \) versus \( \frac{1}{T} \) gives the slope of:
For Arrhenius model of the rate constant \( k = k_0 \cdot e^{-E/RT} \), the graph drawn between \( \ln(k) \) versus \( \frac{1}{T} \) gives the slope of:
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Arrhenius Plot Parameters:
- Plotting \(\ln(k)\) vs \(1/T\) gives a straight line with:
- \(\text{Slope} = -\frac{E}{R}\)
- \(\text{Intercept} = \ln(k_0)\)
- If plotting \(\log_{10}(k)\) vs \(1/T\), the slope becomes \(-\frac{E}{2.303 \cdot R}\).
Updated On: Jul 9, 2026
- \( \frac{E}{R} \)
- \( -\frac{E}{R} \)
- \( \frac{R}{E} \)
- \( -\frac{R}{E} \)
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The Correct Option is B
Solution and Explanation
Concept:
The Arrhenius equation describes the relationship between the reaction rate constant ($k$) and temperature ($T$):
\[
k = k_0 \cdot \exp\left( -\frac{E}{R \cdot T} \right)
\]
To determine the activation energy ($E$) experimentally, the equation can be linearized by taking the natural logarithm ($\ln$) of both sides. This converts the exponential relationship into a linear form that can be plotted as a straight line.
Step 1: Linearizing the Arrhenius equation.
Let us apply the natural logarithm to both sides of the Arrhenius expression: \[ \ln(k) = \ln\left[ k_0 \cdot \exp\left( -\frac{E}{R \cdot T} \right) \right] \] Using the logarithmic identity $\ln(a \cdot b) = \ln(a) + \ln(b)$, we can split the right side into separate terms: \[ \ln(k) = \ln(k_0) + \ln\left[ \exp\left( -\frac{E}{R \cdot T} \right) \right] \] Since the natural logarithm and the exponential function are inverses of each other ($\ln(e^x) = x$), this simplifies to: \[ \ln(k) = \ln(k_0) - \frac{E}{R \cdot T} \] Let us rearrange this equation to group the temperature variable explicitly as an independent term ($1/T$): \[ \ln(k) = \left( -\frac{E}{R} \right) \cdot \left(\frac{1}{T}\right) + \ln(k_0) \]
Step 2: Comparing with the standard straight-line equation.
Let us compare this linearized equation to the standard algebraic equation for a straight line ($y = m \cdot x + c$):
• Dependent variable plotted on the vertical axis (\( y \)): \( y = \ln(k) \)
• Independent variable plotted on the horizontal axis (\( x \)): \( x = \frac{1}{T} \)
• Vertical axis intercept (\( c \)): \( c = \ln(k_0) \)
• Slope of the straight line (\( m \)): \( m = -\frac{E}{R} \) Therefore, a plot of $\ln(k)$ versus $\frac{1}{T}$ yields a straight line with a constant negative slope equal to $-\frac{E}{R}$.
Step 1: Linearizing the Arrhenius equation.
Let us apply the natural logarithm to both sides of the Arrhenius expression: \[ \ln(k) = \ln\left[ k_0 \cdot \exp\left( -\frac{E}{R \cdot T} \right) \right] \] Using the logarithmic identity $\ln(a \cdot b) = \ln(a) + \ln(b)$, we can split the right side into separate terms: \[ \ln(k) = \ln(k_0) + \ln\left[ \exp\left( -\frac{E}{R \cdot T} \right) \right] \] Since the natural logarithm and the exponential function are inverses of each other ($\ln(e^x) = x$), this simplifies to: \[ \ln(k) = \ln(k_0) - \frac{E}{R \cdot T} \] Let us rearrange this equation to group the temperature variable explicitly as an independent term ($1/T$): \[ \ln(k) = \left( -\frac{E}{R} \right) \cdot \left(\frac{1}{T}\right) + \ln(k_0) \]
Step 2: Comparing with the standard straight-line equation.
Let us compare this linearized equation to the standard algebraic equation for a straight line ($y = m \cdot x + c$):
• Dependent variable plotted on the vertical axis (\( y \)): \( y = \ln(k) \)
• Independent variable plotted on the horizontal axis (\( x \)): \( x = \frac{1}{T} \)
• Vertical axis intercept (\( c \)): \( c = \ln(k_0) \)
• Slope of the straight line (\( m \)): \( m = -\frac{E}{R} \) Therefore, a plot of $\ln(k)$ versus $\frac{1}{T}$ yields a straight line with a constant negative slope equal to $-\frac{E}{R}$.
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