If reaction at 500K proceeds at 400 K in presence of catalyst then what is value of Ea ; if catalyst lowers activation barrier by 20kJ/mol???
Asked by Umesh
20th May 2017,
10:04 AM
Answered by Expert
Answer:
Consider Ea and E’a be the energy of activation in presence and absence of catalyst for hydrogenation reaction.
We know,
K = Ae–Ea/RT
So in presence of catalyst,
K1 = Ae–Ea/(R × 500)
And in absence of catalyst,
K2 = Ae–E’a/(R × 400)
Since the two rates are given to be same,
r1 = r2 => K1 = K2
So, Ae–Ea/(R × 500) = Ae–E’a/(R × 400)
Ea/(R × 500) = E’a/(R × 400)
Ea/500 = (Ea – 20)/400 [As, Ea – E’a = 20]
Hence Ea = 100 kJ mol–1
Answered by Expert
20th May 2017,
6:42 PM
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