Answer: The standard Gibbs free energy of the given reaction is 6.84 kJ
Explanation:
For the given chemical equation:
[tex]3A(g)+4B(g)\rightleftharpoons 2C(g)+3D(g)[/tex]
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{(p_C)^2\times (p_D)^3}{(p_A)^3\times (p_B)^4}[/tex]
We are given:
[tex](p_A)_{eq}=5.70atm\\(p_B)_{eq}=4.00atm\\(p_C)_{eq}=4.22atm\\(p_D)_{eq}=5.52atm[/tex]
Putting values in above expression, we get:
[tex]K_p=\frac{(4.22)^2\times (5.52)^3}{(5.70)^3\times (4.00)^4}\\\\K_p=0.0632[/tex]
To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = ?
R = Gas constant = 8.314 J/K mol
T = temperature = [tex]25^oC=[273+25]K=298K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant at 25°C = 0.0632
Putting values in above equation, we get:
[tex]\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.0632)\\\\\Delta G^o=6841.7J=6.84kJ[/tex]
Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ
