Question

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 4.81 atm, PB = 4.26 atm, PC = 5.96 atm, and PD = 5.99 atm.

3A(g) + 3B(g) <-----> C(g) + 3D(g)

What is the standard change in Gibbs free energy of this reaction at 25 °C?

Answer 1

The relation between K eq and Delta Go is as follows:

Delta Go = - RT ln Keq

Here R = gas constant

T = temperature , 25 C or 298 K

Now calculate the value of Kp or Keq as follows:

Kp = product / reactants

For this reaction, Kp = PD^3 X PC / PA^3PB^3

Kp = (5.99 )^3 (5.96 ) / (4.81)^3 (4.26)^3

= 1280.93/ 111.28* 77.31

= 1280.93/ 8603.0568

= 0.1489

Now,

Delta Go = - RT ln Kp

= -8.314 J/molK X 298K X ln 0.1489

= -8.314 J/molK X 298K X (-1.90)

= 4718.8 J/mol

= 4.72 kJ/mol