The standard free energy of activation of a reaction A is 82.1 kJ mol–1 (19.6 kc

Question

The standard free energy of activation of a reaction A is 82.1 kJ mol–1 (19.6 kcal mol–1) at 298 K. Reaction B is one million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ mol–1 (2.39 kcal mol–1) more stable than the reactants.

(a) What is the standard free energy of activation of reaction B?

(b) What is the standard free energy of activation of the reverse of reaction A?

(c) What is the standard free energy of activation of the reverse of reaction B?

Explanation / Answer

use:
K1= A*e^(-Ea1/RT)           ....eqn 1
K2= A*e^(-Ea2/RT)        ...eqn 2

divide eqn 2 by eqn 1
K2/K1 = e^((Ea1-Ea2)/RT)
10^6 = e^((82100-Ea2)/8.314*298)
13.816= (82100-Ea2)/8.314*298
82100-Ea2 = 34229
Ea2= 47871 J/mol
a)
As calculated above Answer is 47871 J/mol = 47.87 J/mol
b)
For reverse of A = Ea1 + 10 KJ/mol = 82.1 KJ/mol + 10 KJ/mol = 92.1 KJ/mol
c)
For reverse of B = Ea2 + 10 KJ/mol = 47.87 KJ/mol + 10 KJ/mol = 57.87 KJ/mol

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