Question

For the system: H2(g) + Cl2(g) ↔ 2HCl(g) ; ΔGo = -190.6 kJ at 25oC. Calculate the equilibrium constant. (be careful of the negative sign)

A.	2.6 x 1033
B.	0.0023
C.	3.9 x 10-34

Answer 1

Ans. Standard Gibb’s free energy, dG⁰ = -RT ln k          - equation 1

                        Where,

                        dG⁰ = Standard Gibb's free energy

                        R = Universal gas constant = 0.0083146 kJ mol^-1 K^-1

T = Temperature in kelvin

k = Equilibrium constant

Given, dG⁰ = -190.6 kJ/mol             ; Temperature, T = 25.0⁰C = 298.15 K

Note: Express dG⁰ in terms of kJ/mol but in kJ alone.

Putting the values in equation 1-

            -190.6 kJ/mol = - (0.0083146 kJ mol^-1 K^-1) x 298.15 K (ln k)

            Or, ln k = 190.6 kJ mol^-1 / 2.47899799 kJ mol^-1

            Or, 2.303 log k = 76.8859

            Or, log k = 76.8859 / 2.303 = 33.38511

            Or, k = antilog 33.38511

            Or, k = 2.427 x 10³³

Hence, equilibrium constant for the reaction - 2.427 x 10³³

Correct option. A. 2.6 x 10³³        (nearest value)