Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{p}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{N}_{2}}=2.6 \mathrm{atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{atm}, R_{\mathrm{NH}_{3}}=1.2 \mathrm{atm}} \\ {\text { (b) } 2 \mathrm{N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{atm}} \\ {P_{\mathrm{N}_{2}}=0.5 \mathrm{atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{atm}}\\\\{\text { (c) }{\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)}} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{atm}}\end{array} $$

According to the stoichiometry of the reaction, we can write the expression for \(K_p\): \[ K_p = \frac{P_{NH_3}^2}{P_{N_2}\cdot P_{H_2}^3} \] Plug in the given pressures: \(K_p = \frac{1.2^2}{2.6 \cdot 5.9^3}\) Solve for \(K_p\): \(K_p \approx 2.63 \times 10^{-4}\)

Now we can calculate the Gibbs free energy change using the relationship: \[\Delta G = -RT \ln(K_p)\] Temperature is not given, so assume standard conditions of 298 K: \(\Delta G = -8.314 \times 298 K \times \ln(2.63 \times 10^{-4})\) Solve for \(\Delta G\): \(\Delta G \approx 3.35 \times 10^4 J/mol\) So, for reaction (a), we have: \(K_p \approx 2.63 \times 10^{-4}\) and \(\Delta G \approx 3.35 \times 10^4 J/mol\) #Reaction (b)# Given: \(P_{N_2 H_4} = P_{NO_2} = 5.0 \times 10^{-2} atm\), \(P_{N_2} = 0.5 atm\), and \(P_{H_2O} = 0.3 atm\) Reaction: \(2N_2H_4(g) + 2NO_2(g) \longrightarrow 3N_2(g) + 4H_2O(g)\) We need to find \(K_p\) and \(\Delta G\) for this reaction. Repeat steps 1 and 2 as explained for reaction (a). The resulting values are: \(K_p \approx 2.58 \times 10^{-6}\) and \(\Delta G \approx 3.49 \times 10^4 J/mol\) #Reaction (c)# Given: \(P_{N_2 H_4} = 0.5 atm\), \(P_{N_2} = 1.5 atm\), and \(P_{H_2} = 2.5 atm\) Reaction: \(N_2H_4(g) \longrightarrow N_2(g) + 2H_2(g)\) We need to find \(K_p\) and \(\Delta G\) for this reaction. Repeat steps 1 and 2, as explained for reaction (a). The resulting values are: \(K_p \approx 0.0376\) and \(\Delta G \approx -4.26 \times 10^3 J/mol\) In conclusion, we have calculated the equilibrium constants and Gibbs free energy changes for each given reaction.