Q.For the reaction, $\mathrm{CH_4(g) + H_2(g) \rightarrow C_2H_6(g)}$, $K_p$ = 3.356 $\times$ 10$^{17}$ Calculate $\Delta G^0$ for the reaction at 25 $^0$C.

[!NOTE]
The equation is transcribed exactly as the textbook prints it -- as printed it is not balanced in carbon. The intended reaction is the hydrogenation $\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)}$ (whose $\Delta G^0$ at 298 K matches the printed answer, and which exercise question 4 xix prints correctly). The calculation below depends only on $K_p$, not on the equation.

For a reaction aA + bB to cC + dD at conditions away from equilibrium, the Gibbs energy change is given by ΔG = ΔG0 + RT ln Q, where ΔG0 is the STANDARD Gibbs energy change (with all reactants and products in their standard states) and Q is the REACTION QUOTIENT -- an expression in the same form as the equilibrium constant, but built from the actual (non-equilibrium) concentrations Qc = [C]^c[D]^d / [A]^a[B]^b or, for gas-phase reactions, partial pressures Qp = (PC^c PD^d) / (PA^a PB^b). When a reaction reaches equilibrium, ΔG becomes exactly zero and Q takes on its special equilibrium value, the equilibrium constant (Kc or Kp); substituting these into the general equation gives 0 = ΔG0 + RT ln K, which rearranges to the key relation connecting thermodynamics and chemical equilibrium: ΔG0 = -RT ln K = -2.303 RT log10 K. This single equation lets the equil …