Question

The standard free energy of activation of a reaction A is 75.1 kJ mol–1 (17.9 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

Arrhenius equation gives the relationship between temperature and the rate constant for a reaction obeyed the following equation.

In this equation, k is the rate constant for the reaction, Z is a proportionality constant that varies from one reaction to another, E_a is the activation energy for the reaction, R is the ideal gas constant in joules per mole kelvin, and T is the temperature in kelvin.

where

R is the gas law constant 1.98 x 10^-3 kcal/mol
&
T is your 298 Kelvin

Rate of reaction A = 1 = Ze^{-17.9 kcal/mol/(1.98 x 10-3 kcal/mol x 298 K)}

Rate of reaction B = 10⁶ = Ze^{-G kcal/mol/(1.98 x 10-3 kcal/mol x 298 K)}

lowering the activation barrier by delta G,

the rate of the reaction is enhanced
by the factor e^dG/RT.

10⁶ = e^{dG/(1.98 x 10-3 kcal/mol x 298 K)}

ln 10⁶ = dG/(1.98 x 10^-3 kcal/mol x 298 K)

13.81 = dG/(1.98 x 10^-3 kcal/mol x 298 K)

dG = 13.81 x (1.98 x 10^-3 kcal/mol x 298 K)

dG = 8.151 kcal/mol

the standard free energy of activation of reaction B = 17.9 k cal/mol - 8.151 kcal/mol = 9.748 kcal/mol