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# The activation energy for the reaction

4.9     The activation energy for the reaction $2HI(g)\rightarrow H_{2}+I_{2}(g)$  is $209.5 KJ mol^{-1}$ at $518 K$.  Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

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We have

Activation energy = 209.5KJ/mol

temperature= 581K

R = 8.314J/mol/K

Now,  the fraction of molecules of reactants having energy equal to or greater than activation energy is given as

$x= e^{-E_{a}/RT}$

taking log both sides we get

$\log x = -\frac{E_{a}}{RT}$

$=\frac{209500Jmol^{-1}}{2.303\times 8.314Jmol^{-1}K^{-1}\times 581}$

= 18.832

x = antilog(18.832)

= 1.471$\times 10^{-19}$

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