Consider the following reaction at \(800 . \mathrm{K}:\) $$\mathrm{N}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{atm}, P_{\mathrm{NF}_{3}}=0.48\) atm. Calculate \(\Delta G^{\circ}\) for the reaction at \(800 .\) K.

Understanding the phenomenon of chemical equilibrium is crucial for grasping the behavior of reactions like the one given in our problem. At chemical equilibrium, the rate at which the reactants turn into products is equal to the rate at which products revert back to reactants. This state doesn't mean the reactants and products are present in equal amounts, but rather that their concentrations remain constant over time.

It's essential to note that chemical equilibrium can be dynamic, where both forward and reverse reactions are still occurring, but the overall effect is no net change in the concentration of reactants and products. This concept is fundamental in predicting how changes in conditions, like temperature or pressure, can shift the equilibrium and alter concentrations.

The equilibrium constant, represented as K, provides a quantitative measure of a reaction's composition at equilibrium. It's a dimensionless number obtained from the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the balanced chemical equation. In the case of gases, pressure is used in place of concentration, giving us a Kp value.

For the given nitrogen-fluorine reaction, the equilibrium constant is expressed in terms of partial pressures, providing a snapshot of the system's position at equilibrium. High Kp values, much greater than 1, suggest product-favored reactions at equilibrium, while low Kp values imply reactant-favored equilibria. Our calculated Kp of 662.22 indicates a large favor towards the formation of NF3 under the provided conditions.

The reaction quotient, Q, is similar in form to the equilibrium constant but applies to systems not necessarily at equilibrium. It uses the same expression as K but with the current concentrations or partial pressures of the reactants and products.

Comparing the reaction quotient, Q, to the equilibrium constant, K, tells us the direction in which the reaction needs to move to reach equilibrium. If Q is less than K, the forward reaction is favored to form more products. If Q is more than K, the reverse reaction is favored to form more reactants. In this problem, the equilibrium state is already established, so Q equals K. However, knowledge of Q is beneficial if the system's conditions change and the reaction shifts out of equilibrium.

The realm of thermodynamics gives us the capacity to predict the spontaneity and feasibility of a chemical reaction through Gibbs free energy change, ΔG. This value takes into account not only the energy levels of the reactants and products (ΔH, enthalpy change) but also the disorder introduced into the system (ΔS, entropy change).

For a process at constant temperature and pressure, the sign of ΔG dictates the spontaneity: a negative ΔG indicates a spontaneous process, while a positive value suggests a non-spontaneous one. In our case, the calculated standard Gibbs free energy change of -30.63 kJ/mol at 800 K signifies that the reaction can occur spontaneously under standard conditions. It's also important to remember that reactions can shift away from these standard states, affecting the ΔG and, hence, the spontaneity of the process.