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

Explanation:

For the given chemical equation:

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

The expression of $K_p$ for the given reaction:

$K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}$

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:

$K_p=\frac{6.90\times 5.90}{8.00\times (5.40)^2}\\\\K_p=0.174$

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:

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

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