Problem 90 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

For reaction (a), we found that \(\frac{Q_p}{K_p} = 0.00219\), so the expression for \(\Delta G\) is \(\Delta G = RT\ln(0.00219)\). For reaction (b), we found that \(\frac{Q_p}{K_p} = 18000\), so the expression for \(\Delta G\) is \(\Delta G = RT\ln(18000)\). For reaction (c), we found that \(\frac{Q_p}{K_p} = 18.75\), so the expression for \(\Delta G\) is \(\Delta G = RT\ln(18.75)\). Without further information on temperature and equilibrium constants, we cannot find the exact values for \(\Delta G\).

Step 1: Reaction (a) - Gibbs Free Energy Calculation

First, we need to use the equation for \(\Delta G\) when the pressure of the components is given: \[\Delta G = RT\ln\left(\frac{Q_p}{K_p}\right)\] Where R is the universal gas constant, T is the temperature, \(Q_p\) is the reaction quotient at the given pressures, and \(K_p\) is the equilibrium constant in terms of pressure. Since we are not given temperature or \(K_p\), we will assume they are constant and only calculate the \(\ln\left(\frac{Q_p}{K_p}\right)\) part. For reaction (a): \[Q_p = \frac{P_{NH_3}^2}{P_{N_2} \cdot P_{H_2}^3}\] Substituting the given pressures: \[Q_p = \frac{(1.2\,\text{atm})^2}{(2.6\,\text{atm}) \cdot (5.9\,\text{atm})^3}\] Now, calculate the value of \(Q_p\): \[Q_p \approx 0.00219\] Thus, for reaction (a): \[\frac{Q_p}{K_p} = 0.00219\] Now, the final expression for \(\Delta G\) of reaction (a) becomes: \[\Delta G = RT\ln(0.00219)\]

Step 2: Reaction (b) - Gibbs Free Energy Calculation

For reaction (b), let's first write the equation for \(Q_p\): \[Q_p = \frac{P_{N_2}^3 \cdot P_{H_2O}^4}{P_{N_2H_4}^2 \cdot P_{NO_2}^2}\] Substituting the given pressures: \[Q_p=\frac{(0.5\,\text{atm})^3 \cdot (0.3\,\text{atm})^4}{(5.0\times10^{-2}\,\text{atm})^2 \cdot (5.0\times10^{-2}\,\text{atm})^2}\] Now, calculate the value of \(Q_p\): \[Q_p \approx 18000\] Thus, for reaction (b): \[\frac{Q_p}{K_p} = 18000\] Now, the final expression for \(\Delta G\) of reaction (b) becomes: \[\Delta G = RT\ln(18000)\]

Step 3: Reaction (c) - Gibbs Free Energy Calculation

For reaction (c), let's first write the equation for \(Q_p\): \[Q_p = \frac{P_{N_2} \cdot P_{H_2}^2}{P_{N_2H_4}}\] Substituting the given pressures: \[Q_p = \frac{(1.5\,\text{atm}) \cdot (2.5\,\text{atm})^2}{(0.5\,\text{atm})}\] Now, calculate the value of \(Q_p\): \[Q_p = 18.75\] Thus, for reaction (c): \[\frac{Q_p}{K_p} = 18.75\] Now, the final expression for \(\Delta G\) of reaction (c) becomes: \[\Delta G = RT\ln(18.75)\] For each reaction, if we were given the temperature T and equilibrium constant \(K_p\), we could now find the exact values for the Gibbs free energy changes.

Key Concepts

Reaction Quotient

The reaction quotient, symbolized as \( Q \), is a key concept when studying chemical reactions and their progression towards equilibrium. It is calculated using the same expression as the equilibrium constant \( K \), but using the initial concentrations or partial pressures of the reactants and products. The formula for \( Q_p \), when dealing with gases, is derived from the law of mass action.

For a generic reaction, \(aA + bB \rightleftharpoons cC + dD\), the reaction quotient \( Q_p \) is determined by: \[ Q_p = \frac{{P_C}^c {P_D}^d}{{P_A}^a {P_B}^b} \] where \(P_X\) represents the partial pressure of the substance \(X\).

If \( Q_p < K_p \), the reaction will proceed in the forward direction to reach equilibrium. Conversely, if \( Q_p > K_p \), the reaction will proceed in the reverse direction. When \( Q_p = K_p \), the reaction is at equilibrium, meaning no net change in the concentration of reactants and products. Calculating \( Q_p \) allows us to predict the direction in which the reaction will proceed under given conditions.

Equilibrium Constant

The equilibrium constant \( K \) is a fundamental concept that gives us insight into the balance point of a chemical reaction. It is a specific value at a given temperature that expresses the ratio of the concentrations of products to reactants at equilibrium.

For gaseous reactions, \( K_p \) is the equilibrium constant expressed in terms of partial pressures. It is calculated similarly to \( Q_p \), but instead of using the initial or non-equilibrium pressures, the equilibrium pressures are used. The formula is the same: \[ K_p = \frac{{P_C}^c{P_D}^d}{{P_A}^a{P_B}^b} \]

The magnitude of \( K_p \) gives us an indication of the position of equilibrium:

• If \( K_p \) is much greater than 1, products are favored at equilibrium.
• If \( K_p \) is much less than 1, reactants are favored.
• If \( K_p \) is approximately 1, neither reactants nor products are particularly favored, implying comparable concentrations of both at equilibrium.

Understanding \( K_p \) allows chemists to predict how a reaction mixture will behave when conditions are altered and to calculate necessary conditions for desired outcomes.

Pressure Calculations

Pressure calculations play a crucial role in determining the behavior of gases in chemical reactions. When dealing with gaseous reactions, partial pressures of the reactants and products need to be considered, as they are directly used to calculate the reaction quotient \( Q_p \) and equilibrium constant \( K_p \).

Partial pressure is the pressure exerted by an individual gas in a mixture of gases and is calculated using Dalton’s Law of Partial Pressures. If you know the total pressure \( P_{total} \) of a system and the mole fraction \( X_i \) of a gas, the partial pressure \( P_i \) of each gas is calculated as:

• \( P_i = X_i \times P_{total} \)

Accurate pressure measurements are essential, especially since they influence \( Q_p \) calculations. For instance, increasing the pressure by decreasing the volume (at constant temperature) might affect the direction in which a reaction proceeds.

By mastering pressure calculations, one gains critical insights into the dynamics and outcomes of high-temperature and -pressure reactions, which are common in industrial processes and laboratory experiments.