Please show all of the steps. Thank you! The standard free energy of activation

Question

Please show all of the steps. Thank you!

The standard free energy of activation of a reaction A is 87.7 kj mol-1 (21.0 kcal mol-1) at 298 K. Reaction 6 is one hundred 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. What is the standard free energy of activation of reaction B? What is the standard free energy of activation of the reverse of reaction A? What is the standard free energy of activation of the reverse of reaction B?

Explanation / Answer

(a) Arrhenius equation: k = A exp(-Ea/RT)

k(B)/k(A) = exp(-(Ea(B) - Ea(A))/RT)

100 x 10^6 = exp(-(Ea(B) - 87700)/(8.314 x 298))

-(Ea(B) - 87700)/(8.314 x 298) = ln(100 x 10^6) = 18.421

Activation energy Ea(B) = 42061 J/mol = 42.1 kJ/mol


(b) Ea(reverse A) = Ea(A) + |DH|

= 87.7 + 10.0 = 97.7 kJ/mol


(c) Ea(reverse B) = Ea(B) + |DH|

= 42.1 + 10.0 = 52.1 kJ/mol

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