Question

Which of the following plot represents the variation of ln k versus \( \frac{1}{T} \) in accordance with the Arrhenius equation?

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is C

The Arrhenius equation is: \[ k = Ae^{-E_a/RT} \]. Here, $k$ represents the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin.
By taking the natural logarithm of both sides, we get: \[ \ln k = \ln A - \frac{E_a}{RT} \].
Rearranging this into the form of a straight line equation ($y = mx + c$) yields: \[ \ln k = -\frac{E_a}{R} \left( \frac{1}{T} \right) + \ln A \].
In this form:
$y = \ln k$
$x = 1/T$
$m = -E_a/R$ (slope)
$c = \ln A$ (y-intercept)
Given that activation energy ($E_a$) and the gas constant ($R$) are positive, the slope ($-E_a/R$) will consistently be negative. Consequently, a plot of $\ln k$ against $1/T$ will display as a straight line with a negative slope, aligning with option (3).