The standard free energy of activation of a reaction A is 77.1 kJ mol–1 (18.4 kc

Question

The standard free energy of activation of a reaction A is 77.1 kJ mol–1 (18.4 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?

Map ro Organic Chemistry Loudon Roberts & Company Publishers presented by Sapling Learning The standard free energy of activation of a reaction A is 77.1 kJ mol- (18.4 kcal mol 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 mol1 (2. than the reactants. 39 kcal mol) more stable (a) What is the standard free energy of activation of reaction B? Number kJ mol (b) What is the standard free energy of activation of the reverse of reaction A? Number kJ mol- (c) What is the standard free energy of activation of the reverse of reaction B? Number kJ mol

Explanation / Answer

a)

Rate B = 10*10^6 Rate A

then

According to Arrhenius, we can relate the rate constants as following:

K = A*exp(-Ea/(RT))

Where:

K = rate constant at Temperature “T”

A = Frequency Factor

E = Activation Energy in J/mol

R = ideal gas constant, 8.314 J/mol-K

T = absolute temperature

Rate B = 10*10^6 Rate A

Kb = 10*10^6 * Ka

Kb/Ka = exp((-Eab - Eaa) /(RT))

(10^8) = exp((-Eab - 77100) /(8.314*298))

ln((10^8)) = (-Eab - 77100) /(8.314*298)

(-Eab - 77100) = ln((10^8))*(8.314*298)

Eab = - ( ln((10^8))*(8.314*298)-77100)

Eab = -31461.43 J/mol = -31.46 kJ/mol

b)

Eareverse = EaA + Eenthalpy

Eareverse =77100 + 10000

Ea-reverse = 87100 J/mol = 87.1 kJ/mol

c)

Ea(B reverse) = Ea(b) + H

Ea(B reverse) = 31.46 + 10 = 41.46 kJ/mol

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