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

Answer: (C)

Answer 2

The Arrhenius equation is given by:
$k = Ae^{-E_a/RT}$
where: $^*$ $k$ is the rate constant; $^*$ $A$ is the pre-exponential factor; $^*$ $E_a$ is the activation energy; $^*$ $R$ is the gas constant; $^*$ $T$ is the temperature in Kelvin.
Taking the natural logarithm of both sides:
$\ln k = \ln A - \frac{E_a}{RT}$
This equation can be rearranged into the form of a straight line equation ($y = mx + c$):
$\ln k = -\frac{E_a}{R} \left( \frac{1}{T} \right) + \ln A$
where:
$y = \ln k$
$x = 1/T$
$m = -E_a/R$ (slope)
$c = \ln A$ (y-intercept)
Since activation energy ($E_a$) and the gas constant ($R$) are always positive, the slope ($-E_a/R$) will always be negative. Therefore, the plot of $\ln k$ versus $1/T$ should be a straight line with a negative slope. This corresponds to option (3).