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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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best_answer

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}

 

Posted by

manish

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