Problem 48 - General Chemistry: Principles and Modern Applications

Question

At \(400^{\circ} \mathrm{C}, \Delta G^{\circ}=-25 \mathrm{kJ}\) for the reaction \(2 \mathrm{Hg}(1)+\) \(\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HgO}(\mathrm{s}) .\) If a sample of \(\mathrm{HgO}(\mathrm{s})\) is heated to \(400^{\circ} \mathrm{C},\) what will be the equilibrium partial pressure of \(\mathrm{O}_{2}(\mathrm{g}) ?\)

Step-by-Step Solution

Answer

The equilibrium partial pressure of the Oxygen gas at 400°C is equal to the calculated value of Kp from Step 1.

Step 1: Convert Gibbs free energy change to the equilibrium constant using the relation between them

The relationship between Gibbs free energy change at standard conditions represented as \( \Delta G^{\circ} \) and the equilibrium constant Kp is given by the equation \( \Delta G^{\circ} = -RT \ln K_{p} \). Given that \( \Delta G^{\circ} = -25 \) kJ, the temperature T = 400°C = 673.15 K (converting Celsius to Kelvin by adding 273.15), and R = 8.3145 J/molK, we need to first convert 25 kJ into Joules by multiplying with 1000 which gives \( \Delta G = -25000 J \). Now we can solve for Kp : \( K_{p} = e^{(- \Delta G^{\circ} / RT )} \) = \( e^{(-(-25000)/ (8.3145*673.15))}.\)

Step 2: Write the equilibrium expression for the given reaction

The equilibrium expression for the reaction, \(2 \Hg(l)+ \O_{2}(g) \rightarrow 2 \HgO(s)\), is represented as \(K_{p}=\frac{ \left[ \O_{2} \right] }{ [\Hg]^{2} },\) where the brackets represent partial pressures of respective elements. Since the product Mercury(II) Oxide \( \HgO(s) \) is a solid, it does not appear in the equilibrium expression.

Step 3: Solving for the unknown partial pressure of Oxygen

Referencing the equilibrium expression, to find the equilibrium partial pressure of \( \O_{2}\), such should be expressed as: \( [\O_{2}] = K_{p} * [\Hg]^{2} \). As no value is given for the partial pressure of \( \Hg \), we will assume the partial pressure of one mole of gas at standard conditions, which is 1 atm. Therefore, \( [\O_{2}] = K_{p} \). Now we substitute the value of Kp we calculated in step 1 to find the equilibrium partial pressure of \( \O_{2} \).

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant (K) is critical in understanding how chemical reactions reach a state of balance called equilibrium. In a closed system, when the rate of the forward reaction equals the rate of the reverse reaction, the concentrations of the reactants and products remain constant over time. This state is described by the equilibrium constant.

The equilibrium constant is a number that provides the ratio of the concentrations of the products to the concentrations of the reactants, raised to the power of their respective coefficients in the balanced chemical equation. For reactions involving gases, partial pressures are used instead of concentrations, and the equilibrium constant is represented as Kp. The equation for Kp takes into account the stoichiometry of the reaction, written as an expression with product partial pressures in the numerator and reactant partial pressures in the denominator. For example, in the case of the reaction given in the exercise, where Mercury (II) Oxide decomposes into mercury and oxygen, we treat the gases involved using their partial pressures to define Kp.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the relationship between chemical reactions and energy changes involving heat. One of the most important concepts in this field is Gibbs free energy, represented by G, which combines enthalpy, entropy, and temperature to predict whether a process will occur spontaneously at constant pressure and temperature.

Gibbs free energy is expressed through the equation \( G = H - TS \), where H represents enthalpy, T temperature, and S entropy. A negative value of \( \Delta G \), or change in Gibbs free energy, implies that the reaction occurs spontaneously. An important aspect of chemical thermodynamics is understanding how \( \Delta G \), at standard conditions (notated as \( \Delta G^\circ \)), is related to the equilibrium constant. This relationship is critical because it links the thermodynamic favorability of the reaction with the position of equilibrium.

Partial Pressure

Partial pressure is a term used in chemistry to describe the pressure exerted by a single type of gas in a mixture of gases. It's directly proportional to the mole fraction of the gas in the mixture and the total pressure exerted by the entire mixture. In the context of chemical reactions involving gases, partial pressure plays a key role in determining the direction and extent of the reaction.

For a reaction at equilibrium involving gases, the partial pressures can be inserted into the equilibrium expression to calculate the equilibrium constant, Kp. This means that partial pressure not only indicates the concentration of a gas in a chemical system but also influences the position of equilibrium in reactions. In the exercise, for instance, the equilibrium partial pressure of oxygen would determine how far the reaction has proceeded to reach equilibrium at a given temperature.

Reaction Quotient

The reaction quotient (Q) is a measure that indicates the relative amounts of products and reactants present during a reaction at a given moment in time. It serves as a predictor for the direction a reaction will shift in order to reach equilibrium. It is similar in form to the equilibrium constant, using the same expression, but relies on initial concentrations or partial pressures rather than those at equilibrium.

If Q is less than K, the forward reaction is favored, and the system will shift to the right to form more products and reach equilibrium. If Q exceeds K, the system has an excess of products, and the reverse reaction is favored until equilibrium is attained. When Q equals K, the system is at equilibrium, and no further shift will occur. By calculating Q and comparing it to K, one can determine the shift required for a reaction mixture to reach equilibrium, a concept that is deeply intertwined with the principles of chemical thermodynamics.