Question

Calculate the pressure of \(\mathrm{O}_{2}\) (in \(\mathrm{atm}\) ) over a sample of \(\mathrm{NiO}\) at \(25^{\circ} \mathrm{C}\) if \(\Delta G^{\circ}=212 \mathrm{~kJ} / \mathrm{mol}\) for the reaction: $$ \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) $$

Short answer

The pressure of \( \mathrm{O}_2 \) is approximately \( 3.35 \times 10^{-75} \text{ atm} \).

Step-by-step solution

Identify Reaction and Given Data

The reaction given is \( \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \). The standard Gibbs free energy change \( \Delta G^{\circ} \) for this reaction is given as 212 kJ/mol. We need to calculate the pressure of \( \mathrm{O}_{2} \) at equilibrium at 25°C.

Write the Expression for the Equilibrium Constant

For the reaction \( \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \), the equilibrium constant \( K_p \) is given by: \( K_p = P_{\mathrm{O}_2}^{1/2} \), where \( P_{\mathrm{O}_2} \) is the partial pressure of \( \mathrm{O}_{2} \).

Relate Gibbs Free Energy Change to Equilibrium Constant

Use the relationship between Gibbs free energy change and equilibrium constant: \[ \Delta G^{\circ} = -RT \ln K_p \] where \( R = 8.314 \text{ J/mol K} \) is the ideal gas constant and \( T = 298 \text{ K} \) is the temperature in Kelvin.

Solve for \( K_p \)

Rearrange the equation to solve for \( K_p \): \[ K_p = e^{-\Delta G^{\circ} / RT} \] Substitute the values: \( \Delta G^{\circ} = 212,000 \text{ J/mol} \), \( R = 8.314 \text{ J/mol K} \), and \( T = 298 \text{ K} \). Calculate \( K_p \): \[ K_p = e^{-(212000) / (8.314 \times 298)} \] which simplifies to \( K_p = e^{-85.619} \).

Calculate \( P_{\mathrm{O}_2} \)

Since \( K_p = P_{\mathrm{O}_2}^{1/2} \), solve for \( P_{\mathrm{O}_2} \): \[ P_{\mathrm{O}_2} = (K_p)^2 \] Calculate \( K_p \) from \( e^{-85.619} \) and then find \( P_{\mathrm{O}_2} \). Calculating \( K_p \) results in \( K_p \approx 5.79 \times 10^{-38} \), thus \( P_{\mathrm{O}_2} = (5.79 \times 10^{-38})^2 \approx 3.35 \times 10^{-75} \text{ atm} \).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_p \), is a numerical representation of the balance between the products and reactants in a chemical reaction at equilibrium. It is crucial in understanding how far a reaction will proceed.
For gas-phase reactions, \( K_p \) is expressed in terms of partial pressures. In our given reaction, \( \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s) + \frac{1}{2} \mathrm{O}_2(g) \), the equilibrium constant involves the partial pressure of \( \mathrm{O}_2 \) as it is the only gaseous component in the system.

• The equation for \( K_p \) is: \( K_p = P_{\mathrm{O}_2}^{1/2} \),
• This takes into account the stoichiometry of the reaction, with respect to the gaseous \( \mathrm{O}_2 \).

The smaller or larger the value of \( K_p \) indicates the extent to which the reaction has occurred or the position of equilibrium, whether it is more towards products or reactants.

Partial Pressure

Partial pressure is a fundamental concept in understanding reactions involving gases. It describes the pressure exerted by a single type of gas in a mixture of gases. In a chemical equilibrium involving gases, partial pressures of the reactants and products determine the equilibrium position.
For the reaction \( \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s) + \frac{1}{2} \mathrm{O}_2(g) \), the partial pressure of \( \mathrm{O}_2 \) is crucial. Since we are asked to find this, we first find the equilibrium constant \( K_p \) and then solve for the pressure using:

• \( P_{\mathrm{O}_2} = (K_p)^2 \).

Partial pressures help us understand how much \( \mathrm{O}_2 \) is present at equilibrium, and is essential for predicting the behavior of reactions in closed systems.

Equilibrium Reaction

An equilibrium reaction is one where the forward and reverse reactions occur at the same rate. This results in no net change in the amounts of reactants and products over time, creating a dynamic balance.
In our context with \( \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s) + \frac{1}{2} \mathrm{O}_2(g) \), both the decomposition of \( \mathrm{NiO} \) and the recombination to form \( \mathrm{NiO} \) are in balance.

• The system reaches a state where the concentrations (or pressures) of reactants and products are constant.
• This balance is quantitatively described by the equilibrium constant \( K_p \).

Understanding equilibrium reactions is crucial, as it allows us to predict the conditions needed to either drive the reaction to completion or maintain a certain state of balance.

Ideal Gas Constant

The ideal gas constant, denoted as \( R \), is a crucial factor in calculations involving gases in chemical thermodynamics. Its value is commonly \( 8.314 \text{ J/mol K} \).
In the context of equilibrium calculations, \( R \) is used in the relation between the Gibbs free energy change and the equilibrium constant:\[- \Delta G^{\circ} = RT \ln K_p.\]

• This relationship helps determine the extent of reaction at a given temperature and allows us to connect thermodynamic properties to observable quantities such as pressure.
• Without \( R \), it would be impossible to relate these two important aspects of chemical reactions effectively.

The ideal gas constant supports our calculations by providing a link between the amount of energy needed to change states and how this translates into pressures and equilibria in gases.