Question

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 8.00 atm, PB = 5.40 atm, PC = 6.90 atm, and PD = 5.90 atm.

A(g) + 2B(g) -----> C(g) + D(g)

What is the standard change in Gibbs free energy of this reaction at 25 degrees Celsius

Answer 1

Answer: The standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol

Step-by-step explanation:

For the given chemical equation:

`A(g)+2B(g)\rightleftharpoons C(g)+D(g)`

The expression of
`K_p` for the given reaction:

`K_p=(p_C* p_D)/(p_A* (p_B)^2)`

We are given:

`p_A=8.00atm\\p_B=5.40atm\\p_C=6.90atm\\p_D=5.90atm`

Putting values in above equation, we get:

`K_p=(6.90* 5.90)/(8.00* (5.40)^2)\\\\K_p=0.174`

To calculate the standard Gibbs free energy, we use the relation:

`\Delta G^o=-RT\ln K_p`

where,

`\Delta G^o` = standard Gibbs free energy

R = Gas constant =
`8.314J/K mol`

T = temperature =
`25^oC=[25+273]K=298K`

`K_p` = equilibrium constant in terms of partial pressure = 0.174

Putting values in above equation, we get:

`\Delta G^o=-(8.314J/Kmol)* 298K* \ln (0.174)\\\\\Delta G^o=4332.5J/mol=4.33kJ/mol`

Hence, the standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol