Question

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

Short answer

For the given reactions, we have calculated the equilibrium constants (\(K_p\)) and Gibbs free energy changes (\(\Delta G\)) as follows: - 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 \(\Delta G \approx -117 \text{ kJ/mol}\)

Step-by-step solution

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: \[K_p = \frac{P_\text{products}^\text{coefficients}}{P_\text{reactants}^\text{coefficients}}\]

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) ###

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: \[K_p \approx 7.33 \times 10^{-9}\]

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) ###

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: \[K_p \approx 2.00 \times 10^8\]

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) ###

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: \[K_p \approx 971\]

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 \(\Delta G \approx -117 \text{ kJ/mol}\)

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_p\) in the context of gases, gives us a snapshot of a chemical reaction's balance between products and reactants at equilibrium. This constant is a valuable indicator of a reaction's position: whether it favors the products or reactants more once equilibrium is reached. To calculate \(K_p\), use the formula:\[K_p = \frac{P_\text{products}^\text{coefficients}}{P_\text{reactants}^\text{coefficients}}\]Here, \(P\) denotes the partial pressure of each species involved, with the corresponding coefficients representing their stoichiometric coefficients from the balanced equation.
You raise each \(P\) to the power of its coefficient because at equilibrium, these pressures give a ratio that remains constant under constant temperature.
Understanding this helps in predicting how changes in conditions might shift the equilibrium.
For instance, in reaction (a) involving ammonia formation, we calculated \(K_p = 7.33 \times 10^{-9}\). This small value indicates a strong preference for the reactants at equilibrium.
Knowing \(K_p\) values allows chemists to make informed decisions about reaction conditions to either increase yield or speed.

Gibbs Free Energy

Gibbs free energy, denoted by \( \Delta G \), is a central concept in thermodynamics and provides insight into a reaction's spontaneity. It combines enthalpy, entropy, and temperature factors into a single value. The relationship is given by:\[ \Delta G = \Delta G^\circ + RT\ln{Q} \]\(Q\) is the reaction quotient, which, like \(K_p\), is calculated from the product and reactant pressures but at any point during the reaction (not just equilibrium).
\( \Delta G^\circ \) is the standard Gibbs free energy change, derived under standard conditions (1 atm and 25°C).
A negative \( \Delta G \) indicates a reaction that is spontaneous under the given conditions.
In the calculation for reaction (a), adaptations were made to account for non-standard conditions using pressure values.
\( \Delta G \approx -21.7 \text{ kJ/mol} \) shows that even under current conditions, the reaction slightly favors product formation despite the small \(K_p\).
This dual understanding of \( \Delta G \) and \(K_p\) helps determine the feasibility and yield of any industrial or laboratory reaction process.

Thermodynamic Calculations

Thermodynamic calculations take a methodical approach to quantify the favorability and extent of chemical reactions. They are based on principles like energy conservation and dispersal, suited to calculate both \(K_p\) and \( \Delta G \). When starting calculations, one must:

• Identify all chemical reactions involved.
• Gather necessary data, including temperature (usually in Kelvin) and pressure data from reliable sources like stated appendices.
• Apply the formulas for \(K_p\) and \( \Delta G \).
• Understand how changes in these calculations reveal useful outcomes, like shifts in reaction favorability.

These calculations are iterative in some cases because different reactions influence each other's states through shared reactants or products.
Reactions proceed to minimize \( \Delta G \), leading toward equilibrium.
In reaction (c), as shown, the calculations reveal significant \(K_p\) and \( \Delta G \), demonstrating shifts that occur when reactants convert to stable products under evolving pressures.
These thorough thermodynamic calculations are indispensable for chemical engineers and scientists when predicting reaction behavior in response to changes in conditions like pressure and temperature.