Question

According to Arrhenius equation rate constant k is equal to ${\displaystyle A{e}^{\frac{-Ea}{RT}}}$. Which of the following options represents the graph of ln k vs ${\displaystyle \frac{1}{T}}$?

Options

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Answer 1

Explanation:

To arrhenius equation ${\displaystyle k=A{e}^{\frac{-Ea}{RT}}}$ 

Taking log on both sides in ${\displaystyle k}$ = In ${\displaystyle A.e^{-\frac{E_a}{RT}}}$

In ${\displaystyle k}$ = In ${\displaystyle A-\frac{E_a}{RT}}$

In ${\displaystyle k=-\frac{-E_a}{R}\times \frac{1}{T}}$ + In A

y = mx + c

This equation can be related to equation of straight line.

From the graph, it is very clear that slope of the plot = ${\displaystyle \frac{-E_a}{R}}$ and intercept = In A.