Question

From Arrhenius law, \(\textbf{k} = \textbf{k}_0\textbf{e}^{-\textbf{E/RT}}\), a plot of \(\ln(\textbf{k})\) versus 1/T gives straight line with a slope of _______

• A. E/R
• B. R/E
• C. \(-E/R\)
• D. \(-R/E\)

Hint:

An Arrhenius plot has a negative slope of \(-E/R\).
Because the slope is negative, a higher slope in magnitude (steeper line) indicates a reaction with a higher activation energy, meaning its rate constant is highly sensitive to temperature changes.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the slope of the straight line obtained when plotting the natural logarithm of the rate constant (\(\ln(k)\)) against the reciprocal of the absolute temperature (\(1/T\)) using the Arrhenius equation.

Step 2: Key Formula or Approach:
The Arrhenius equation describes the temperature dependence of reaction rate constants:
\[ k = k_0 e^{-E/RT} \] where:
\(k\) is the reaction rate constant.
\(k_0\) is the pre-exponential factor (frequency factor).
\(E\) is the activation energy of the reaction.
\(R\) is the universal gas constant.
\(T\) is the absolute temperature in Kelvin.
To find the slope of the plot, we take the natural logarithm of both sides of the Arrhenius equation and arrange it into the standard straight-line equation form \(y = mx + c\).

Step 3: Detailed Explanation:
Let us apply the natural logarithm (\(\ln\)) to both sides of the Arrhenius equation:
\[ \ln(k) = \ln\left(k_0 e^{-E/RT}\right) \] Using the logarithmic identity \(\ln(A \cdot B) = \ln(A) + \ln(B)\):
\[ \ln(k) = \ln(k_0) + \ln\left(e^{-E/RT}\right) \] Since \(\ln(e^x) = x\), the equation simplifies to:
\[ \ln(k) = \ln(k_0) - \frac{E}{RT} \] Now, we rearrange the terms to isolate the variable plotted on the horizontal axis, which is \(1/T\):
\[ \ln(k) = \left( -\frac{E}{R} \right) \left( \frac{1}{T} \right) + \ln(k_0) \] By comparing this directly to the equation of a straight line:
\[ y = m \cdot x + c \] where:
\(y = \ln(k)\) (vertical axis variable)
\(x = 1/T\) (horizontal axis variable)
\(m = \text{slope of the line}\)
\(c = y\text{-intercept}\)
Comparing the corresponding terms, we find:
\[ \text{Slope } (m) = -\frac{E}{R} \] \[ y\text{-intercept } (c) = \ln(k_0) \] Thus, the slope of the line is \(-E/R\), which is negative, meaning the rate constant increases as temperature increases (as \(1/T\) decreases).

Step 4: Final Answer
The slope is \(-E/R\), which corresponds to option (C).