Consider the following reaction at 298 K: \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) \(K_{P}=0.142\) What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.

To understand how chemical reactions reach a state of balance, it's essential to grasp the concept of the equilibrium constant. The equilibrium constant, symbolized as K, is a numerical value that reflects the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their respective stoichiometric coefficients in the balanced equation.

For the reaction at hand, \( \mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \), the equilibrium constant under conditions of constant pressure is given as \(K_P = 0.142\). This tells us that, at equilibrium, the products (in this case \( \mathrm{NO}_2 \) gas) will not hugely dominate over the reactants (\( \mathrm{N}_2\mathrm{O}_4 \)) because \(K_P\) isn't a very large number. A large \(K_P\) would indicate a greater extent of reaction, with more products being formed. However, \(K_P\) merely reflects the state of the system at equilibrium; it does not indicate how fast the equilibrium will be reached.

Delving into the concept of Gibbs free energy is crucial when analyzing chemical reactions and their spontaneity. Gibbs free energy, often abbreviated as \(G\), is a thermodynamic quantity that combines enthalpy, temperature, and entropy to predict the favorability of a process. The standard free energy change, \(\Delta G^\circ\), is particularly helpful since it tells us whether a reaction will occur spontaneously under standard conditions.

Using the formula \(\Delta G^\circ = -RT\ln(K)\), we can calculate this value as was demonstrated in the solution steps for the given equilibrium constant \(K_P\) and temperature in Kelvin. A negative \(\Delta G^\circ\) implies a spontaneous reaction, while a positive value suggests non-spontaneity. In our exercise, \(\Delta G^\circ\) is positive, indicating that the formation of \(\mathrm{NO}_2\) from \(\mathrm{N}_2\mathrm{O}_4\) is not spontaneously favored under standard state conditions at 298 K.

Chemical equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of the reactants and products. It isn't a static state but rather a dynamic balance where reactants and products continuously interconvert, maintaining stable concentrations over time.

Our reaction \( \mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \) will tend toward equilibrium starting from the standard state. This doesn't mean all the \(\mathrm{N}_2\mathrm{O}_4\) will be converted into \(\mathrm{NO}_2\); rather, it will adjust concentrations until the ratio of the partial pressures corresponds to the given equilibrium constant. Importantly, reaching equilibrium does not depend on the starting amounts of reactants or products, but the final equilibrium state will always correspond to the same equilibrium constant at a given temperature.

Reaction spontaneity is a term used to describe whether a chemical process will occur without the need for an external input of energy. Spontaneous reactions are those that have the potential to proceed on their own once initiated. It's important to understand that spontaneous does not necessarily mean fast; it simply means energetically favorable.

For the reaction \( \mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \), we’ve calculated a positive value of \(\Delta G^\circ\). This suggests that the reaction, under standard conditions at 298 K, is not spontaneous—an input of energy would be required to drive the reaction towards producing more \(\mathrm{NO}_2\). Outside of standard state or initial conditions, spontaneity can shift, and other factors such as changes in temperature, pressure, or concentration can affect whether a reaction will be spontaneous.