Answer: The standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol
Explanation:
For the given chemical equation:
[tex]A(g)+2B(g)\rightleftharpoons C(g)+D(g)[/tex]
The expression of [tex]K_p[/tex] for the given reaction:
[tex]K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}[/tex]
We are given:
[tex]p_A=8.00atm\\p_B=5.40atm\\p_C=6.90atm\\p_D=5.90atm[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{6.90\times 5.90}{8.00\times (5.40)^2}\\\\K_p=0.174[/tex]
To calculate the standard Gibbs free energy, we use the relation:
[tex]\Delta G^o=-RT\ln K_p[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = [tex]25^oC=[25+273]K=298K[/tex]
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = 0.174
Putting values in above equation, we get:
[tex]\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.174)\\\\\Delta G^o=4332.5J/mol=4.33kJ/mol[/tex]
Hence, the standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol
