Answer
(a) -24.6$\frac{KJ}{mol}$
(b) -1.33$\frac{KJ}{mol}$
Work Step by Step
(a)The Gibbs free energy change of a reaction ($\Delta$$G_{R}$) at a certained temperature (T) is: $\Delta$$G_{R}$=$\Delta$$G_{R}^{o}$+RTln($Q_p$).
At equilibrium, $\Delta$$G_{R}$=$\Delta$$G_{R}^{o}$+RTln($K_p$)=0. So, the standard Gibbs free energy change of the reaction ($\Delta$$G_{R}^{o}$):
$\Delta$$G_{R}^{o}$=-RTln($K_p$)=-8.31$\frac{J}{mol K}$$\times$2000K$\times$ln(4.40)$\approx$-24624$\frac{J}{mol}$=-24.6$\frac{KJ}{mol}$
*** Notice standard Gibbs free energy change of the reaction ($\Delta$$G_{R}^{o}$) is under the standard condition, in which the partial pressure of all gaseous components is 1 atm. ***
(b)For this reaction, at a state with given partial pressures $P_{H_{2}}$, $P_{CO_{2}}$, $P_{H_{2}O}$ & $P_{CO}$, the Gibbs free energy change of the reaction ($\Delta$$G_{R}$) under the same temperature (T) is:
$\Delta$$G_{R}$=$\Delta$$G_{R}^{o}$+RTln($Q_p$)
=$\Delta$$G_{R}^{o}$+RTln($\frac{(P_{H_{2}O}) (P_{CO})}{(P_{H_{2}}) (P_{CO_{2}})}$)
=-24624$\frac{J}{mol}$+8.31$\frac{J}{mol K}$$\times$2000K$\times$ln($\frac{(0.66)(1.20)}{(0.25)(0.78)}$)
$\approx$-1330$\frac{J}{mol}$=-1.33$\frac{KJ}{mol}$
