Problem 64 - Chemistry: Principles and Reactions

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 \(\Delta G_{i}^{\circ}\) for \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Question: Calculate (a) the standard Gibbs free energy change (ΔG°) for the given reaction and (b) the standard Gibbs free energy of formation (ΔGᵢ°) for N2O at 25°C, given the following information: Reaction: N2O + NO2 → 3NO K = 4.4 × 10⁻¹⁹ ΔG°(NO) = 87.5 kJ/mol ΔG°(NO2) = 52.1 kJ/mol Answer: (a) ΔG° for the reaction at 25°C = 2.67 × 10² kJ/mol (b) ΔGᵢ° for N2O at 25°C = 8.4 kJ/mol

Step 1: Part (a) - Calculate \(\Delta G^{\circ}\) for the reaction at 25°C

To calculate \(\Delta G^{\circ}\) for the reaction, we will use the formula connecting the equilibrium constant (K) with the standard Gibbs free energy change: $$ \Delta G^{\circ} = -RT \ln K $$ where \(R\) is the gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin (298 K) and \(K\) is the equilibrium constant given as \(4.4 \times 10^{-19}\). Plug in the values and solve for \(\Delta G^{\circ}\): $$ \Delta G^{\circ} = - (8.314 \,\text{J/mol K})(298 \,\text{K}) \ln (4.4 \times 10^{-19}) $$ $$ \Delta G^{\circ} \approx 2.67 \times 10^{2} \,\text{kJ/mol} $$

Step 2: Part (b) - Calculate \(\Delta G_{i}^{\circ}\) for N2O at 25°C

To calculate the standard Gibbs free energy of formation for N2O, we will use the equation for the standard Gibbs free energy change in terms of the standard Gibbs free energies of formation of the products and reactants: $$ \Delta G^{\circ} = \sum_\text{products}(\text{nG}) - \sum_\text{reactants}(\text{nG}) $$ For the given reaction: $$ \Delta G^{\circ} = 3 \Delta G_{\mathrm{NO}}^{\circ} - (\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} + \Delta G_{\mathrm{NO}_{2}}^{\circ}) $$ We are given the standard Gibbs free energy of formation for NO2 and NO: $$ \Delta G_{\mathrm{NO}_{2}}^{\circ} = 52.1 \,\text{kJ/mol} $$ $$ \Delta G_{\mathrm{NO}}^{\circ} = 87.5 \,\text{kJ/mol} $$ Insert the values and the calculated \(\Delta G^{\circ}\) from part (a): $$ 2.67 \times 10^{2} \,\text{kJ/mol} = 3 (87.5 \,\text{kJ/mol}) - (\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} + 52.1 \,\text{kJ/mol}) $$ Now, solve for \(\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ}\): $$ \Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} = 3 (87.5 \,\text{kJ/mol}) - 2.67 \times 10^{2} \,\text{kJ/mol} - 52.1 \,\text{kJ/mol} $$ $$ \Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} \approx 8.4 \,\text{kJ/mol} $$

Key Concepts

Understanding Equilibrium Constant

The equilibrium constant, often represented as \( K \), is a fundamental concept in chemical reactions. It describes the ratio of the concentrations of products to reactants at equilibrium. When a reaction reaches equilibrium, the forward and reverse reaction rates are equal. This balance means the concentrations of the reactants and products remain constant over time, but it's essential to note that they do not have to be equal to each other.

In the expression for equilibrium constant:

• For a reaction \( aA + bB \leftrightarrow cC + dD \), the equilibrium constant \( K \) is given by \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).
• The brackets \([ ]\) denote concentration in terms of molarity (moles per liter).

This ratio is temperature dependent. As seen with your given reaction, even though the \( K \) value is extremely small (\( 4.4 \times 10^{-19} \)), it tells us that at equilibrium, the concentration of products (NO) is very low compared to reactants (N₂O and NO₂). Such a tiny number indicates that the reaction does not spontaneously form many products under the given conditions.

Thermodynamics and Gibbs Free Energy

Thermodynamics is the study of energy transformations. Gibbs free energy, denoted by \( \Delta G \), is a crucial thermodynamic quantity that indicates the spontaneity of a chemical reaction. The equation \( \Delta G^{\circ} = -RT \ln K \) directly relates Gibbs free energy to the equilibrium constant \( K \). This equation uses the gas constant \( R \) (8.314 J/mol K) and the absolute temperature \( T \).

The relationship can be summarized as follows:

• If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous and will proceed in the forward direction.
• If \( \Delta G^{\circ} = 0 \), the system is at equilibrium.
• If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous and tends to go in the reverse direction.

For the given reaction at 25°C (which is \( 298 \) Kelvin), the positive \( \Delta G^{\circ} \approx 2.67 \times 10^{2} \,\text{kJ/mol} \) indicates that at standard conditions, the reaction is not spontaneous. This aligns with the very low equilibrium constant, showing that energy is needed for the formation of significant amounts of product.

The Dynamics of Chemical Reactions

Chemical reactions involve the transformation of reactants into products. Understanding a reaction's dynamics, such as rate and extent, depends heavily on concepts like reaction mechanisms, energy changes, and ultimately the equilibrium constant.

Our exercise involves a seemingly simple reaction: \( \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \rightarrow 3 \mathrm{NO}(g) \). Though it looks straightforward, reactions often occur in complex steps that influence how and when equilibrium is reached.

• The rates at which reactants are converted to products depend on factors like temperature, pressure, and catalysts.
• The energy changes associated with reactions are explained using Gibbs free energy, which considers both enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)).
• The ability to predict reaction spontaneity and direction heavily relies on thermodynamics, as well as the calculated \( \Delta G \).

In summary, while chemical reactions are driven towards equilibrium, understanding them fully requires an interplay of kinetics, thermodynamics, and molecular behavior. This deeper insight is what guides scientists to optimize conditions in various industrial and laboratory processes.