Question

Using the data in Appendix \(\mathrm{C}\) and given the pressures listed, calculate \(K_{\rho}\) and \(\Delta G\) for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) \(R_{\mathrm{N}_{2}}=2.6 \mathrm{~atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{~atm}, P_{\mathrm{NH}_{5}}=1.2 \mathrm{~atm}\) (b) \(2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=\mathrm{PNo}_{1}=5.0 \times 10^{-2} \mathrm{~atm}\). \(P_{\mathrm{N}_{2}}=0.5 \mathrm{~atm}, P_{\mathrm{H}_{2} O}=0.3 \mathrm{~atm}\) (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g})\) \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{~atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{~atm}, P_{\mathrm{H}_{1}}=2.5 \mathrm{~atm}\)

Short answer

(a) For the reaction $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$, we find \(K_{\rho} \approx 5.25 \times 10^{-5}\) and \(\Delta G \approx 5.45 \times 10^{4}\,\mathrm{J/mol}\). (b) For the reaction \(2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\), we find \(K_{\rho} \approx 2.16 \times 10^{11}\) and \(\Delta G \approx -1.24 \times 10^{5}\,\mathrm{J/mol}\). (c) For the reaction $\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g})$, we find \(K_{\rho} \approx 4.50\) and \(\Delta G \approx -2.34 \times 10^{3}\,\mathrm{J/mol}\).

Step-by-step solution

(a) Calculate \(K_{\rho}\) for (a)

First, we need to calculate the reaction quotient \(K_\rho\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) using the given pressures. The formula for \(K_{\rho}\) is given by: \[K_{\rho} = \frac{P_{\mathrm{NH}_3}^2}{(P_{\mathrm{N}_{2}})(P_{\mathrm{H}_{2}})^3}\] Substitute the given pressures into the formula: \[K_{\rho} = \frac{(1.2\,\mathrm{atm})^2}{(2.6\,\mathrm{atm})(5.9\,\mathrm{atm})^3}\] Now, evaluate \(K_{\rho}\).

(a) Calculate \(\Delta G\) for (a)

To calculate \(\Delta G\) for reaction (a) we will use the Gibbs-Helmholtz equation: \[\Delta G = -RT\ln K_{\rho}\] Given that the data is from Appendix C, assume room temperature, \(T=298.15\,\mathrm{K}\) and the ideal gas constant, \(R=8.314\,\mathrm{J/(mol\cdot K)}\). Now, substitute the calculated \(K_{\rho}\) and given values into the equation and determine \(\Delta G\).

(b) Calculate \(K_{\rho}\) for (b)

Similar to (a), we first calculate the reaction quotient \(K_\rho\) for the reaction \(2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) using the given pressures. The formula for \(K_{\rho}\) is: \[K_{\rho} = \frac{(P_{\mathrm{N}_2})^3(P_{\mathrm{H_2O}})^4}{(P_{\mathrm{N}_{2} \mathrm{H}_{4}})^2(P_{\mathrm{NO}_{2}})^2}\] Substitute the given pressures into the formula and evaluate \(K_{\rho}\).

(b) Calculate \(\Delta G\) for (b)

Now, calculate \(\Delta G\) for reaction (b) using the Gibbs-Helmholtz equation with the calculated \(K_{\rho}\), \(T=298.15\,\mathrm{K}\), and \(R=8.314\,\mathrm{J/(mol\cdot K)}\). Substitute the values into the equation and solve for \(\Delta G\).

(c) Calculate \(K_{\rho}\) for (c)

For reaction (c), $\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g})\(, calculate the reaction quotient \)K_\rho\( using the given pressures. The formula for \)K_{\rho}$ is: \[K_{\rho} = \frac{(P_{\mathrm{N}_2})(P_{\mathrm{H_2}})^2}{(P_{\mathrm{N}_{2} \mathrm{H}_{4}})}\] Substitute the given pressures into the formula and evaluate \(K_{\rho}\).

(c) Calculate \(\Delta G\) for (c)

Lastly, calculate \(\Delta G\) for reaction (c) using the Gibbs-Helmholtz equation with the calculated \(K_{\rho}\), \(T=298.15\,\mathrm{K}\), and \(R=8.314\,\mathrm{J/(mol\cdot K)}\). Substitute the values into the equation and solve for \(\Delta G\).

Key concepts

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a crucial part of understanding chemical equilibrium. It provides a snapshot of a reaction at any given time, not necessarily at equilibrium. The reaction quotient is determined using the same formula as the equilibrium constant \(K\), but the concentrations or pressures of the reactants and products are those at a specific moment. This helps us predict the direction in which a reaction will proceed to reach equilibrium.

For a general reaction \(aA + bB \longrightarrow cC + dD\), the reaction quotient \(Q\) is calculated as follows:

• \(Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

In the exercise, you see different reaction quotients calculated for reactions such as \(\mathrm{N}_2 + 3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3\). By inserting the provided pressures into these equations, \(Q\) can be directly found. This calculation is valuable as it indicates whether a system is at equilibrium (\(Q = K\)), not started yet (\(Q < K\)), or has proceeded beyond equilibrium (\(Q > K\)). Thus, \(Q\) is extremely useful for understanding and predicting chemical behavior.

Gibbs Free Energy

Gibbs Free Energy, represented by \(\Delta G\), is a thermodynamic quantity that combines enthalpy and entropy into a value that can predict the spontaneity of a reaction. It indicates the maximum amount of reversible work that can be done by a thermodynamic system at constant temperature and pressure. The formula used for this calculation in relation to the reaction quotient is given by:

• \(\Delta G = -RT\ln K\)

Where \(R\) is the ideal gas constant and \(T\) is the temperature in Kelvin.

In the context of this exercise, \(\Delta G\) helps in understanding whether a chemical reaction will occur spontaneously. A negative \(\Delta G\) implies that the reaction is spontaneous under the given conditions, meaning it can proceed towards equilibrium without any external energy input. Conversely, a positive \(\Delta G\) shows that the reaction is non-spontaneous, requiring external energy to proceed. Thus, \(\Delta G\) is a powerful tool for chemists to assess both the feasibility and directionality of chemical reactions.

Ideal Gas Constant

The ideal gas constant \(R\) is an essential component of the ideal gas law and thermodynamic equations, including those related to chemical equilibria and Gibbs Free Energy. It serves as a universal conversion factor between different units of pressure, volume, and temperature. The value of \(R\) is consistent across various equations used in chemistry and physics, making it crucial for calculations involving gases.

• Value: \(R = 8.314 \mathrm{J/(mol \cdot K)}\)

In the exercise, \(R\) is utilized for calculating \(\Delta G\) through the Gibbs-Helmholtz equation. It links the reaction quotient \(K_\rho\) and temperature \(T\) to determine the change in Gibbs Free Energy. Understanding the ideal gas constant's role in these calculations provides students with insight into real-world applications of equilibrium and thermodynamics, highlighting how theoretical concepts underpin practical chemical analysis.