Write euilibrium constant for the each : (a) \(\quad \mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(\mathrm{~g})}\) (b) \(\quad \mathrm{KClO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{KCl}_{(\mathrm{s})}+(3 / 2) \mathrm{O}_{2(\mathrm{~g})}\) (c) \(\mathrm{CaC}_{2(\mathrm{~s})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{CaCO}_{3(\mathrm{~s})}+2 \mathrm{CO}_{2(\mathrm{~g})}\) (d) \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) (e)Fe \(^{3+}{ }_{\text {(aq })}+\mathrm{SCN}_{\text {(aq.) }}^{*}=\mathrm{Fe}(\mathrm{SCN})^{2+}{ }_{\text {(aq })}\) (f) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} \rightleftharpoons \mathrm{CuSO}_{4(\mathrm{~s})}+5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{v})}\)

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, not because the reactions have stopped, but because they are proceeding at the same rate.

Understanding chemical equilibrium involves grasping that it's a dynamic process, where the amounts of reactants and products don't change. The equilibrium constant (\( K \) ) is the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their respective coefficients from the balanced chemical equation.

In the original exercise, the equilibrium constants for various reactions were defined. An equilibrium constant is specific to a particular reaction at a given temperature. For instance, reaction (a) given by \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \) has an equilibrium constant \( K_{(a)} = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \), which provides valuable insights into the relative quantities of \( \text{N}_2\text{O}_4 \) and \( \text{NO}_2 \) present at equilibrium.

The reaction quotient (\( Q \) ) is a measure that indicates whether a system is at equilibrium and, if not, in which direction it will proceed to reach equilibrium. It is calculated in the same way as the equilibrium constant, but for any given point in time during a reaction, not just at equilibrium.

By comparing \( Q \) with the equilibrium constant (\( K \)), one can deduce whether the reaction will proceed forwards (\( QK \) ), which implies that there are excess products, and the system will favor the reactants. When \( Q \) equals \( K \) , the system is at equilibrium, and there are no net changes in the concentrations of reactants and products.

\( Q \) is especially useful when we are analyzing changes in a reaction that has not yet reached equilibrium, allowing us to predict which way the reaction will shift to achieve a state of balance.

Le Chatelier's Principle provides a qualitative prediction of how a system at equilibrium responds to changes in concentration, temperature, or pressure. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

For instance, increasing the concentration of reactants will shift the equilibrium position to produce more products. Similarly, if we remove products, the system will tend to produce more products to re-establish equilibrium. This principle is crucial in industrial chemical processes where maximizing product yield is desirable.

In the context of the exercise, if we consider reaction (d) \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \) and increase the concentration of \( \text{N}_2 \) or \( \text{H}_2 \), the principle predicts that more ammonia (\( \text{NH}_3 \) ) will be produced to restore equilibrium. Le Chatelier's Principle thus helps us understand and control the conditions to direct the reaction to yield more of a desired product.