CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For the system A(g)B(g),Hfor the forward reaction is 33KJ/mol (Note: H=E in this case). What will be the equilibrium constant Kat 300 K. If the activation energies Ef & Eb are in the ratio 20:31, Calculate Ef and Eb at this temperature? (Assume that the pre-exponential factor is the same for the forward and backward reactions)

A
K=5.572×105
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
K=5.572×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
K=557.2×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A K=5.572×105
EfEb=ΔH(1)
Let Ef be 20x & Eb be 31x.
where x is the common factor.
20x-31x = -33
-11x = -33
x = 3
Now, Ef = 60 KJ/mol & Eb = 93 KJ/mol
Using arrhenius equation.
Kf=Ae60000(300)R
Kb=Ae93000300R
By definition of equilibrium constant Keq=KfKb
Keq=e60000300R+93000300R
e33000300Re110/8.314
It is 5.572×105

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