wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The Activation Energy For The Reaction 2HI(g)H2+I2(g)Is209.5Mol-1at581K.Calculate The Fraction Of Molecules Of Reactants Having Energy Equal To Or Greater Than The Activation Energy.


Open in App
Solution

Solution :

Step 1: Given data:

The Arrhenius Equation is given by,

k=Ae-EaRT

Here in this equation, Ea is the activation energy, and the term, k=e-EaRT is the fraction of molecules of reactants having energy equal to or greater than the activation energy.

Step 2: Formula applied:

Therefore,

e-EaRT=e-2095008.314×581=e-43.4

Step 3: Calculating the Fraction Of Molecules :

Now, calculating the values of the Fraction Of Molecules

= 1antilog(43.4)

=1.47×10-19

Hence The Fraction Of Molecules Of Reactants Having Energy Equal To Or Greater Than The Activation Energy are 1.47×10-19.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Thermochemistry
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon