What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$, where $k$ is the rate constant of a reaction at temperature $T$ ?

• A

  $\frac{-E_a}{2.303 R}$
• Correct answer
  B

  $\frac{-E_a}{R}$
• C

  $\frac{E_a}{R}$
• D

  $\frac{R}{E_a}$

Solution

Arrhenius equation is $k=A e^{-E_a / R T}$ On taking log both sides, $\ln k=\ln A-\frac{E_a}{R T}$ or $\quad \ln k=-\frac{E_a}{R T}+\ln A$ Since, the reaction, $y=m x+c$ when a graph is plotted between $\ln k$ and $\frac{1}{T}$ a straight line is obtained. The slope of the straight line is equal to $-\frac{E_a}{R}$.

Hence, the correct answer is option: 'B'.