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The activation energy for the reaction 2HI(g) → H2 + 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? - CBSE (Science) Class 12 - Chemistry

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Question

The activation energy for the reaction 2HI(g) → H2 + I2(gis 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 = 209500 J 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 = e^((-E_a)/"RT")`

`=>"In  x" = - (E_a)/"RT"`

`=>log x = -E_a/(2.303RT)`

`=>log x = - (209500"J mol"^-1)/(2.303xx8.314 JK^(-1)xx581)` = -18.8323

Now, x =Antilog (18.8323)

= 1.471 x 10-19

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Solution The activation energy for the reaction 2HI(g) → H2 + 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? Concept: Temperature Dependence of the Rate of a Reaction.
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