The standard free energy of activation of a reaction A is 88.2 kJ mol–1 (21.1 kc

Question

The standard free energy of activation of a reaction A is 88.2 kJ mol–1 (21.1 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?

Explanation / Answer

Qualitatively, Standard free energy of activation of reactant B must be < 88.2 KJ/mol at 298 K

standard free energy of reverse reaction is ...88.2+10 = 98.2 KJ/mol

Because Ea reactant - Ea product = H

So Ea product = H - Ea reactant

Standard free energy of reverse reaction of B << 88.2 KJ/mol , because a faster reaction has lower Eact than a slower reaction .

As per Arrhenius , k = A e^(-Ea/RT)

log k = log A - Ea / 2.303RT

kog k2/k1 = Ea1/Ea2

We assumed that A is same for reaction A and reaction B but it can be false . Unfortunately in given problem A values are not specified

log 10^7 = 88.2/Ea2 ( ten million times = 10^)

so Ea2 = 88.2 /7 =   12.6 KJ/mole

b) Eact = 88.2 -10 = 78.2

c) Eact = 12.6-10 = 2.6

Further calculations are similar as above

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