Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{\mathrm{p}}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad R_{\mathrm{N}_{2}}=263.4 \mathrm{kPa}, P_{\mathrm{H}_{2}}=597.8 \mathrm{kPa}, P_{\mathrm{NH}_{3}}=101.3 \mathrm{kPa} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.07 \mathrm{kPa}\) $$ \begin{array}{l} \quad R_{\mathrm{N}_{2}}=50.7 \mathrm{kPa}, P_{\mathrm{H}_{2} \mathrm{O}}=30.4 \mathrm{kPa} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ P_{\mathrm{N}_{2} \mathrm{H}_{4}}=101.3 \mathrm{kPa}, P_{\mathrm{N}_{2}}=152.0 \mathrm{kPa}, P_{\mathrm{H}_{2}}=253.3 \mathrm{kPa} \end{array} $$

Step by step solution

01

General equation for Kp

For a reaction, the equilibrium constant \(K_p\) is equal to the product of the partial pressures of the products raised to their stoichiometric coefficients divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients: \[K_p = \frac{P_\text{products}^\text{coefficients}}{P_\text{reactants}^\text{coefficients}}\]

02

General equation for ΔG

The Gibbs free energy change of a reaction at non-standard conditions, \(\Delta G\), is related to the standard Gibbs free energy change, \(\Delta G^\circ\), the gas constant, \(R\), the temperature, \(T\), and the reaction quotient, \(Q\), as follows: \[\Delta G = \Delta G^\circ + RT\ln{Q}\] Where \(Q = \frac{P_\text{products}^\text{coefficients}}{P_\text{reactants}^\text{coefficients}}\). #Step 2: Calculate Kp and ΔG for each reaction# ### Reaction (a) ###

03

Calculate Kp for reaction (a)

Using the given pressures for reactants and products, we can calculate \(K_p\) for reaction (a) as follows: \[K_p = \frac{(P_{\mathrm{NH}_3})^2}{P_{\mathrm{N}_2}(P_{\mathrm{H}_2})^3} = \frac{(101.3 \text{ kPa})^2}{(263.4 \text{ kPa})(597.8 \text{ kPa})^3}\] After calculating, we get: \[K_p \approx 7.33 \times 10^{-9}\]

04

Calculate ΔG for reaction (a)

To calculate \(\Delta G\) for reaction (a), we first need to find \(\Delta G^\circ\) from Appendix C. For the formation of NH3, we have \(\Delta G^\circ = -16.4 \text{ kJ/mol}\). Now, we can calculate \(\Delta G\) using the relationship: \[\Delta G = \Delta G^\circ + RT\ln{Q}\] Since our \(Q \approx K_p\) in this case, we get: \[\Delta G = -16.4 \text{ kJ/mol} + RT \ln{(7.33 \times 10^{-9})}\]. We can assume the temperature in this problem is room temperature, so we assume \(T = 298 \text{ K}\) and use the value of the gas constant in kJ/mol K: \(R = 8.314 \times 10^{-3} \text{ kJ/mol K}\). Finally, we get: \[\Delta G \approx -21.7 \text{ kJ/mol}\] ### Reaction (b) ###

05

Calculate Kp for reaction (b)

Using the given pressures for reactants and products, we can calculate the \(K_p\) for reaction (b) as follows: \[K_p = \frac{(P_{\mathrm{N}_2})^3(P_{\mathrm{H}_2 \mathrm{O}})^4}{(P_{\mathrm{N}_2 \mathrm{H}_4})^2(P_{\mathrm{NO}_2})^2} = \frac{(50.7\text{ kPa})^3(30.4\text{ kPa})^4}{(5.07\text{ kPa})^2(5.07\text{ kPa})^2}\] After calculating, we get: \[K_p \approx 2.00 \times 10^8\]

06

Calculate ΔG for reaction (b)

We need the \(\Delta G^\circ\) values for each compound involved in the reaction. From Appendix C: - For N2H4, \(\Delta G^\circ = 95.4 \text{ kJ/mol}\) - For NO2, \(\Delta G^\circ = 51.3 \text{ kJ/mol}\) - For N2, \(\Delta G^\circ = 0 \text{ kJ/mol}\) - For H2O, \(\Delta G^\circ = -228.6 \text{ kJ/mol}\) We calculate \(\Delta G^\circ\) for the overall reaction as: \[\Delta G^\circ = 3(0) + 4(-228.6) - 2(95.4) - 2(51.3) \approx -939.2 \text{ kJ/mol}\] Now, using the relationship \(\Delta G = \Delta G^\circ + RT\ln{Q}\), we find: \[\Delta G \approx -939.2 \text{ kJ/mol} + RT \ln{(2.00 \times 10^8)}\] After calculating with \(T = 298 \text{ K}\) and \(R = 8.314 \times 10^{-3} \text{ kJ/mol K}\), we get: \[\Delta G \approx -1026 \text{ kJ/mol}\] ### Reaction (c) ###

07

Calculate Kp for reaction (c)

Using the given pressures for reactants and products, we can calculate the \(K_p\) for reaction (c) as follows: \[K_p = \frac{(P_{\mathrm{N}_2})(P_{\mathrm{H}_2})^2}{(P_{\mathrm{N}_2 \mathrm{H}_4})} = \frac{(152.0 \text{ kPa})(253.3 \text{ kPa})^2}{(101.3 \text{ kPa})}\] After calculating, we get: \[K_p \approx 971\]

08

Calculate ΔG for reaction (c)

We already found the \(\Delta G^\circ\) values for N2H4 and N2 from reaction (b). We can use these values to find \(\Delta G\) for reaction (c) using the relationship \(\Delta G = \Delta G^\circ + RT\ln{Q}\): \[\Delta G = -95.4 \text{ kJ/mol} + RT \ln{(971)}\] After calculating with \(T = 298 \text{ K}\) and \(R = 8.314 \times 10^{-3} \text{ kJ/mol K}\), we get: \[\Delta G \approx -117 \text{ kJ/mol}\] After all calculations, the final values are: - Reaction (a): \(K_p \approx 7.33 \times 10^{-9}\) and \(\Delta G \approx -21.7 \text{ kJ/mol}\) - Reaction (b): \(K_p \approx 2.00 \times 10^8\) and \(\Delta G \approx -1026 \text{ kJ/mol}\) - Reaction (c): \(K_p \approx 971\) and \(\Delta G \approx -117 \text{ kJ/mol}\)

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