Problem 53 - General Chemistry: Principles and Modern Applications

Question

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

Step-by-Step Solution

Answer

For part (a), upon solving, we will find that the partial pressure of oxygen at equilibrium is extremely low. For part (b), to obtain significant quantities of oxygen, Lavoisier likely increased the temperature or decreased the pressure to shift the reaction equilibrium toward the products.

Step 1: Formulate the Equation for the Equilibrium Constant \(K_p\)

Express the equilibrium constant in terms of partial pressures. For the given reaction \(\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\), the equilibrium constant \(K_p\) is given by \(K_p = P_{\mathrm{O}_{2}(\mathrm{g})}^{1/2}\), where \(P_{\mathrm{O}_{2}(\mathrm{g})}\) is the partial pressure of oxygen.

Step 2: Calculate \(K_p\) Using Gibbs Free Energy Change Equation

From thermodynamics, we have \(\Delta G^\circ = -RT \ln(K_p)\) where \(\Delta G^\circ\) is the standard Gibbs free energy change, \(R\) is the universal gas constant, \(T\) is temperature in Kelvin, and \(K_p\) is the equilibrium constant. Given \(\Delta G^\circ=+58.54 \mathrm{kJ/mol} = +58.54 \times 10^3 \mathrm{J/mol}\), \(R = 8.314 \mathrm{J/mol.K}\), and \(T = 298 \mathrm{K}\) (corresponding to \(25^{\circ} \mathrm{C}\)), we can solve for \(K_p\).

Step 3: Calculate the Partial Pressure of \(\mathrm{O}_{2}\)

From Step 2, we will get a value for \(K_p\). Since \(K_p = P_{\mathrm{O}_{2}(\mathrm{g})}^{1/2}\), we can find \(P_{\mathrm{O}_{2}(\mathrm{g})}\) by squaring \(K_p\). This will confirm if the partial pressure of oxygen at equilibrium is very low.

Step 4: Discuss the Conditions for Obtaining Significant Quantities of Oxygen

Part (b) requires understanding of the effect of pressure and temperature on the reaction. In order to obtain significant amounts of oxygen from the decomposition of \(\mathrm{HgO}\), Lavoisier would have needed to manipulate these conditions. If \(K_p\) is small, this means the equilibrium lies to the left, favoring the reactants. Hence, to push the reaction to the right, either increase the temperature or decrease the pressure. Specific conditions would depend on practical constraints.

Key Concepts

Gibbs free energy

Gibbs free energy is a thermodynamic quantity that measures the maximum reversible work that can be performed by a thermodynamic system at constant temperature and pressure. It's represented by the symbol \( \Delta G \). A reaction with a negative \( \Delta G \) will proceed spontaneously, meaning it will naturally occur without external energy input. In the case of positive \( \Delta G \), like our decomposition of mercury(II) oxide reaction, the process is non-spontaneous, meaning it requires external energy to proceed.
Understanding \( \Delta G \) is crucial for predicting how a reaction will behave under certain conditions. Here, our \( \Delta G^{\circ} = +58.54 \mathrm{kJ/mol} \) indicates that at standard conditions (25°C, 1 atm), the decomposition of mercury(II) oxide into mercury and oxygen gas doesn't happen by itself.
This value plays a direct role in calculating the equilibrium constant \( K_p \), linking \( \Delta G \) and the condition of the system at equilibrium.

Equilibrium constant

The equilibrium constant \( K_p \) quantifies the relationship between the products and reactants of a reaction at equilibrium. For reactions involving gases, this constant is usually expressed in terms of partial pressures. In our decomposition reaction of mercury(II) oxide, \( K_p \) is calculated as the square root of the partial pressure of \( \mathrm{O}_{2} \).

• \( K_p = P_{\mathrm{O}_{2}(\mathrm{g})}^{1/2} \)

To find this constant, we use the formula involving Gibbs free energy: \( \Delta G^{\circ} = -RT \ln(K_p) \). Substituting the values of \( \Delta G^{\circ}, R, \) and \( T \), we can solve for \( K_p \).
For our specific case, the calculated \( K_p \) is very small, suggesting that the equilibrium vastly favors the reactants over the products. This reflects the low partial pressure of oxygen, meaning the production of \( \mathrm{O}_{2} \) is not favored under standard conditions without extra intervention.

Thermodynamics

Thermodynamics is the study of energy transformations and is vital in understanding chemical reactions and processes. It helps us grasp how different factors such as temperature, pressure, and energy affect the reaction system's state.
For the mercury(II) oxide decomposition, thermodynamics provides us frameworks and formulas like Gibbs free energy and the equilibrium constant to analyze the reaction's behavior. At the heart of this study is the law of conservation of mass, stating that matter is neither created nor destroyed. This principle forms the basis for balancing chemical equations and understanding how reactions proceed.
Temperature is a critical thermodynamic variable. It dictates energy changes in reactions as described by enthalpy \( \Delta H \), and it's combined with entropy \( \Delta S \) to calculate \( \Delta G \). Our reaction shows high \( \Delta H^{\circ} \), meaning it's endothermic. Such reactions need heat to advance, often manipulated during experiments or industrial processes to steer reactions in the desired direction.

Decomposition reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. These reactions often require energy input in the form of heat, light, or electricity.
In our context, mercury(II) oxide decomposes into liquid mercury and oxygen gas when sufficient energy is provided. This is represented by the equation: \( \mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l) + \frac{1}{2} \mathrm{O}_{2}(g) \).
Decomposition reactions, such as the one studied by Lavoisier, are critical in understanding more profound chemical principles. He demonstrated the conservation of mass during this process, showing that the total mass of the products equals the original mass of the reactants.
These reactions have practical applications in various fields, such as metallurgy, where metals are extracted from their ores. In the lab, manipulating external conditions like heat or pressure helps direct the reaction's course, useful in many chemical manufacturing processes.