Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of ±1 in all quantities): a. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{atm}\) b. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=200 \mathrm{atm}, P_{\mathrm{H}_{2}}=600 \mathrm{atm}\) \(P_{\mathrm{NH}_{3}}=200 \mathrm{atm}\)

The equation for calculating Gibbs free energy change (ΔG) in a reaction is: \[\Delta G = \Delta G^\circ + RT \ln Q\] where ΔG° is standard Gibbs free energy of formation, R is the universal gas constant, T is the temperature, and Q is the reaction quotient.

The equilibrium constant (K) is related to the standard Gibbs free energy of formation (ΔG°) by the following equation: \[K = e^{-\frac{\Delta G^\circ}{RT}}\] However, in this problem, we aren't given values for ΔG°, thus we cannot determine K directly.

The reaction quotient (Q) is calculated by using the partial pressures of the reactants and products. For this reaction: \[Q = \frac{P_{NH_3}^2}{P_{N_2} \cdot P_{H_2}^3}\] Let's calculate Q for both sets of conditions: a) T = 298 K, PN2 = PH2 = 200 atm, PNH3 = 50 atm \[Q_a = \frac{50^2}{200 \cdot 200^3}\] b) T = 298 K, PN2 = 200 atm, PH2 = 600 atm, PNH3 = 200 atm \[Q_b = \frac{200^2}{200 \cdot 600^3}\]

Since we don't have exact values for K or ΔG°, we will at least identify whether the reaction will be spontaneous or not spontaneous in each scenario: ΔG = ΔG° + RT ln Q a) If Q < K, then ln(Q) < 0, and ΔG < ΔG° (reaction will be spontaneous in the forward direction) b) If Q > K, then ln(Q) > 0, and ΔG > ΔG° (reaction will be non-spontaneous in the forward direction) Plug in the calculated values of the Q into the ΔG equation: a) ΔG = ΔG° + (8.314 J/mol·K)(298 K) ln(\(\frac{50^2)}{200 \cdot 200^3}\)) Using the condition for spontaneity, we can conclude if the reaction is spontaneous or non-spontaneous for case 'a' without knowing the exact values. b) ΔG = ΔG° + (8.314 J/mol·K)(298 K) ln(\(\frac{200^2}{200 \cdot 600^3}\)) Similarly, using the condition for spontaneity, we can conclude if the reaction is spontaneous or non-spontaneous for case 'b'. Due to the lack of essential information such as the standard Gibbs free energy of formation (ΔG°) and the equilibrium constant (K), we are unable to calculate the exact values for ΔG in both scenarios. However, by calculating the reaction quotient (Q) and comparing it with the equilibrium constant (K), we are still able to determine the spontaneity of the reaction.