Question

Consider the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) \quad K=4.4 \times 10^{-19} $$ (a) Calculate \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\).

Short answer

Question: Calculate the following: (a) The Gibbs free energy change for the reaction N2O(g) + NO2(g) → 3NO(g) at 25°C, given that the equilibrium constant, K, is 4.4 × 10^(-19); (b) The Gibbs free energy of formation for N2O at 25°C, given that the Gibbs free energy of formation for NO is 86.6 kJ/mol and the Gibbs free energy of formation for NO2 is 51.3 kJ/mol. Answer: (a) The Gibbs free energy change for the reaction at 25°C is approximately 108,139 J/mol; (b) The Gibbs free energy of formation for N2O at 25°C is approximately 208.5 kJ/mol.

Step-by-step solution

(Step 1: Write the expression relating ΔG° and K)

The relationship between the Gibbs free energy change of reaction (ΔG°) and the equilibrium constant (K) is given by: $$ \Delta G^\circ = -RT\ln{K} $$ Where R is the gas constant (\(8.314 J\, mol^{-1}\, K^{-1}\)), T is the temperature in Kelvin, and K is the equilibrium constant.

(Step 2: Convert the temperature to Kelvin)

The given temperature value is in Celsius. Convert it to Kelvin: $$ T = 25^{\circ}C + 273.15 = 298.15\:K $$

(Step 3: Calculate ΔG° for the reaction)

Substitute the values of R, T, and K into the expression for ΔG° and calculate the result: $$ \Delta G^\circ = - (8.314\: J\, mol^{-1}\, K^{-1})(298.15\:K) \ln{(4.4\times 10^{-19})} $$ Calculate the value of ΔG°: $$ \Delta G^\circ \approx 108139 J/mol $$

(Step 4: Calculate ΔG° for the formation of N2O)

For the reaction: $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3\: \mathrm{NO}(g) $$ The Gibbs free energy of formation for N2O can be determined from the Gibbs free energy of formation of the other species in the reaction: $$ \Delta G_{\mathrm{f}}^{\circ}(\mathrm{N}_2 \mathrm{O}) = [\Delta G_{\mathrm{f}}^{\circ}(3\: \mathrm{NO})] - [\Delta G_{\mathrm{f}}^{\circ}(\mathrm{NO}_2)] $$ The values of ΔG° can be found in standard tables: $$ \Delta G_{\mathrm{f}}^{\circ}(\mathrm{NO}) = 86.6\;\text{kJ/mol} $$ $$ \Delta G_{\mathrm{f}}^{\circ}(\mathrm{NO}_2) = 51.3\;\text{kJ/mol} $$ Substitute the values and calculate ΔG° for the formation of N2O: $$ \Delta G_{\mathrm{f}}^{\circ}(\mathrm{N}_2 \mathrm{O}) = [(3\cdot86.6)\;\text{kJ/mol}] - [51.3\;\text{kJ/mol}] $$ Calculate the value of ΔG° for the formation of N2O: $$ \Delta G_{\mathrm{f}}^{\circ}(\mathrm{N}_2 \mathrm{O}) \approx 208.5\; \text{kJ/mol} $$ So, the answers to the questions are: a) The Gibbs free energy change for the reaction at \(25^{\circ} \mathrm{C}\) is approximately 108,139 J/mol; b) The Gibbs free energy of formation for N2O at \(25^{\circ} \mathrm{C}\) is approximately 208.5 kJ/mol.

Key concepts

Equilibrium Constant

In the realm of chemical reactions, the equilibrium constant (K) is a fundamental concept demonstrating the ratio of concentrations of products to reactants at equilibrium. It quantifies the extent to which a reaction proceeds under a given set of conditions, essentially reflecting the balance point where the rates of the forward and reverse reactions are equal.

For the reaction \[\mathrm{N}_{2}O(g) + \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)\], K is given as \(4.4 \times 10^{-19}\), indicating a reaction heavily favoring the reactants at equilibrium. This small value suggests that under standard conditions, very little of the reactants convert into the product.

Understanding K is crucial for predicting the direction a reaction will proceed and is intrinsically linked to the concept of Gibbs free energy, instrumental for assessing the spontaneity of chemical processes.

Chemical Thermodynamics

The heart of predicting and understanding the flow of energy during chemical reactions lies in chemical thermodynamics. It deals with the relationship between heat, work, temperature, and chemical changes, involving laws of thermodynamics that describe how these quantities behave. One of the key functions in this field is Gibbs free energy (\(\Delta G\)), which provides pivotal insight into the feasibility and spontaneity of a chemical reaction.

The link between the equilibrium constant and Gibbs free energy is captured by the equation \[\Delta G^\circ = -RT\ln{K}\], where \(\Delta G^\circ\) is the change in Gibbs free energy under standard conditions, R is the universal gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. If \(\Delta G\) is negative, the process is spontaneous and will proceed without external energy input. However, a positive value indicates non-spontaneity, often requiring additional energy to proceed.

Educational exercises typically involve calculating \(\Delta G^\circ\) to assess reaction spontaneity under standard conditions or other scenarios by manipulating this fundamental relationship.

Standard State Conditions

To streamline the study of chemical reactions and provide a common basis for comparison, chemists use standard state conditions. These are predefined conditions of temperature and pressure, typically set at 25°C (298.15 K) and 1 bar (100 kPa) of pressure, under which the properties of substances are measured and reported.

Under these conditions, the Gibbs free energy of formation (\(\Delta G_{\mathrm{f}}^\circ\)) refers to the change in Gibbs free energy when 1 mole of a compound is formed from its elements in their standard states. This value is a crucial piece of the puzzle in solving for unknowns in thermodynamics problems. For instance, solving for the \(\Delta G_{\mathrm{f}}^\circ\) of \(\mathrm{N}_{2}O(g)\) involves using the Gibbs free energies of formation for the substances involved in the reaction under standard state conditions.

Grasping the significance of standard state conditions and how they affect thermodynamic quantities is vital for students not only to solve textbook problems but also to understand the broader context of the physical and chemical behavior of substances.