Question

The standard free energy of activation of a reaction A is 81.9 kJ mol–1 (19.6 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. (a) What is the standard free energy of activation of reaction B?

Answer 1

54.9 kJ/mol

Step-by-step explanation:

The relation between the activation energy (Ea) and the rate constant (k) is given by the Arrhenius equation.

`k=A.e^(-Ea/RT)`

where,

A is a collision factor

R is the ideal gas constant

T is the absolute temperature

Reaction B is one million times faster than reaction A at the same temperature. So
`k_(B)=10^(6) k_(A)`.

Then,

`k_(B)=10^(6) k_(A)\\A.e^{-Ea_(B)/RT}=10^(6)A.e^{-Ea_(A)/RT}\\e^{-Ea_(B)/RT}=10^(6)e^{-Ea_(A)/RT}\\ln(e^{-Ea_(B)/RT})=ln(10^(6)e^{-Ea_(A)/RT})\\(-Ea_(B))/(RT) =ln10^(6) -(Ea_(A))/(RT) \\Ea_(B)=(ln10^(6) -(Ea_(A))/(RT)).(-RT)=(ln10^(6)-(89.1kJ/mol)/((8.314* 10^(-3) kJ/mol.K).298K) ).(-8.314* 10^(-3) (kJ)/(mol.K).298K )=54.9kJ/mol`