Problem 98 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

For each reaction, we can find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) by first calculating the reaction quotients: Reaction (a): \(Q_p = \frac{ (1.2)^2 }{ (2.6) \cdot (5.9)^3 }\) Reaction (b): \(Q_p = \frac{ (0.5)^3 \cdot (0.3)^4 }{ (5.0 \times 10^{-2})^2 \cdot (5.0 \times 10^{-2})^2 }\) Reaction (c): \(Q_p = \frac{ (1.5) \cdot (2.5)^2 }{ (0.5) }\) Then, we can use the equation \(\Delta G = \Delta G^{\circ}+RT\ln Q_p\) to solve for \(\Delta G^{\circ}\) using the values provided in Appendix C and assuming T = 298 K, R = 8.314 J/mol · K.

Step 1: Identify given information

List all given data and unknowns.

Step 2: Determine relevant principles

Identify applicable chemical laws.

Step 3: Set up and solve

Apply equations and solve.

Step 4: State the answer

The answer is: For each reaction, we can find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) by first calculating the reaction quotients: Reaction (a): \(Q_p = \frac{ (1.2)^2 }{ (2.6) \cdot (5.9)^3 }\) Reaction (b): \(Q_p = \frac{ (0.5)^3 \cdot (0.3)^4 }{ (5.0 \times 10^{-2})^2 \cdot (5.0 \times 10^

Key Concepts

Reaction Quotient

The reaction quotient (\textbf{Q}) plays a crucial role in understanding the progress of a reaction at a given moment in time. It is a snapshot of a reaction's ratio of products to reactants, each raised to the power of their stoichiometric coefficients. The equation for a reaction in the gas phase, at pressures P, is:

\[\[\begin{align*}Q_p = \frac{{P_{\text{{products}}}^\text{{stoichiometric coefficient}} }}{{ P_{\text{{reactants}}}^\text{{stoichiometric coefficient}} }}\end{align*}\]\]
Where \textbf{P} represents the partial pressures of the reactants and products. The values of Q can help predict whether a reaction will proceed forward or reverse to achieve chemical equilibrium. If Q is less than the equilibrium constant (\textbf{K}), the reaction tends to produce more products. Conversely, if Q is greater than K, the reaction favors the formation of reactants.

Standard State

The standard state of a substance is a reference point used to define its thermodynamic properties under specified conditions of pressure and temperature. For gases, the standard state is typically 1 atmosphere (atm) of pressure and usually a temperature of 298 K (25°C), unless otherwise specified. The standard Gibbs free energy change (\textbf{\[\[\begin{align*}circle\end{align*}\]\]}), denoted by the superscript °, indicates the amount of energy change when reactants transform into products at standard state conditions. Understanding standard state is vital because it sets a baseline from which we can understand how variations in conditions like pressure and temperature affect the free energy and directionality of a chemical reaction.

Chemical Equilibrium

Chemical equilibrium is a state of balance in which the forward and reverse reactions occur at equal rates, so the concentrations of the reactants and products remain constant over time. At equilibrium, the reaction quotient (\textbf{Q}) equals the equilibrium constant (\textbf{K}), and there is no net change in the Gibbs free energy of the system.\[\[\begin{align*}circle\end{align*}\]\]Understanding this concept is vital because it helps to predict the behavior of reactions under different conditions and is the basis for optimizing reaction conditions in industrial processes. Moreover, the concept of equilibrium does not imply that reactants and products are in equal concentrations, but rather that their rates of formation are equivalent.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. Each gas in the mixture behaves as if it alone occupies the entire volume of the mixture, and its partial pressure is a measure of its concentration in the gas phase. It's a direct way to express the amount of each gas involved in a reaction. In chemical equilibrium and reaction quotient calculations, the partial pressures are used to represent the concentration of gaseous reactants and products. It's important for predicting how changes in pressure can shift equilibrium and alter the Gibbs free energy change of a reaction.

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. In chemistry, it's essential for understanding how energy changes during chemical reactions. The Gibbs free energy, a thermodynamic potential, predicts whether a reaction will happen spontaneously under constant pressure and temperature. The sign of the Gibbs free energy change (\textbf{\[\[\begin{align*}circle\end{align*}\]\]}) determines the spontaneity of a reaction: if negative, the reaction is spontaneous; if positive, it requires external energy to proceed. This allows chemists to assess the feasibility of reactions and to manipulate conditions to favor the production of desired products.