The standard free energy of activation of a reaction A is 82.7 kJ mol–1 (19.8 kc

Question

The standard free energy of activation of a reaction A is 82.7 kJ mol–1 (19.8 kcal mol–1) at 298 K. Reaction B is ten 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?

I asked this before and was given the wrong answer. I know that B is 92.7. But I need a and c. I know the answers are not 8.27 and 18.27 becuase I put them down as answers and was told that they were wrong. Any help???

Explanation / Answer

As there it is given that the product is 10KJ more stable than the reactant that means the energy of B will be 10KJ less to A as it is more stable

So Activation Energy of B = 82.7+10= 92.7 KJ/mol

a) Standard free energy of activation for reaction B = 92.7KJ/mol

b) The standard activation energy for reverse of A = -82.7 KJ/mol

c)The standard activation energy for reverse of B = -92.7 KJ/mol

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