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Question
The activation energy for the reaction \[\ce{2 HI_{(g)} -> H2_{(g)} + I2_{(g)}}\] is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Solution
In the given case:
Ea = 209.5 kJ mol−1
T = 581 K
R = 8.314 JK−1 mol−1
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
x = `"n"/"N"`
x = `e^((-E_a)/"RT")`
In x = `- (E_a)/"RT"`
or, log x = `-E_a/(2.303 "RT")`
or, log x = `-(209.5 xx 10^3 "J mol"^-1)/(2.303 xx 8.314 "JK"^-1 "mol"^-1 xx 581 "K")`
= −18.8323
x = antilog (−18.8323)
= antilog 19.1677
= 1.471 × 10−19
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