The standard free energy of activation of a reaction A is 73.9kJ/mol-1 (17.7 kca

Question

The standard free energy of activation of a reaction A is 73.9kJ/mol-1 (17.7 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.

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

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

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

Explanation / Answer

ln(k2/k1)=(Ea2-Ea1)/RT

or ln(10^6)=(73.9-x)/(8.31*298)

or x=39.687 kJ/mol

2)Ea=-73.9-10

=-83.9 kJ/mol

3)Eb=-39.687-10

=49.687 kJ/mol

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