According to Arrhenius equation rate constant k is equal to $A . e^{- E_{a} / R T .}$

Which of the following options represents the graph of ln k vs $\frac{1}{T}$ ?

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Solution

According to Arhenius equation, $k = A . e^{- E_{a} / R T .}$

Taking log on both side In $k = I n (A . e^{-} \frac{E_{a}}{R T})$
$I n k = I n A - \frac{E_{a}}{R T}$
$I n k = - \frac{- E_{a}}{R} \times \frac{1}{T} + I n A$
$y = m x + c$

This equation can be related to equation of straight line as shown above.

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

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Similar questions

In the equation k = A . e^-E_a^/RT , A is known as

K = A e^{- E a / R T}

k = A e^{- E_{a} / R T}

(k)

(A)

k = A e^{- E_{a} / R T}

A

k = A e^{- E a / R T}

E_{a}

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