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Q.

A graph of lnK Vs $\frac{1}{T}$ is given:

Where K is rate constant $(in \sec^{- 1})$ ,T is temperature (in K)
Then fraction of molecules having energy equal to or greater than threshold energy is

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a

$e^{20 / T}$

b

$e^{20 / R T}$

c

$e^{- 20 / RT}$

d

$e^{- 20 / T}$

answer is D.

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Detailed Solution

$\ln K = \ln A - \frac{E_{a}}{RT}$
Slope $= \frac{- E_{a}}{R} = - 20$
Fraction of molecules $= e^{- Ea / RT} = e^{- 20 / T}$

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