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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 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
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Solution
Given: Energy of activation (Ea) = 209.5 kJ mol−1
Temperature (T) = 581 K
R = 8.314 JK−1 mol−1
Now, the fraction of molecules of reactants having energy equal to or greater than the activation energy is given as:
x = `n/N`
= `e^((-E_a)/(RT))`
ln x = `- (E_a)/(RT)`
or, log x = `-E_a/(2.303 RT)`
= `-(209.5 xx 10^3)/(2.303 xx 8.314 xx 581)`
log x = −18.8323
x = antilog (−18.8323)
= 1.471 × 10−19
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