Question

Using standard Gibbs energy of formation values given in the table, calculate the equilibrium constant K of the reaction

Cl2_(g)+2NO_(g)⇌2NOCl_(g)

The standard Gibbs energy change of the reaction represents the drive the reaction has under standard conditions to move toward equilibrium from point A to point X in the diagram.

Substance	ΔfG∘ (kJ mol−1)
Cl2(g)	0
NO(g)	86.71
NOCl(g)	66.30

Answer 1

where,

= stoichiometric coefficients of products

= stoichiometric coefficients of reactants

Given,

of Cl₂(g) = 0 kJ mol^-1

of NO(g) = 86.71 kJ mol^-1

of NOCl(g) = 66.30 kJ mol^-1

Thus,

or, = (2 x 66.30 kJ mol^-1) - (2 x 86.71 kJ mol^-1)

                    = - 40.82 kJ/mol

                    = - 40.82 x 1000 J/mol       ( 1 kJ = 1000 J)

                    = - 40820 J/mol

Now,

= - RTlnK

where,

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K    (standard condition)

Thus,

- 40820 J/mol = - 8.314 J/mol.K x 298 K x lnK

or, lnK = 16.5

or, K = 14,300,538

         = 1.43 x 10⁷

Hence, the equilibrium constant for the reaction = 1.43 x 10⁷