The equilibrium constant of the reaction; \(\mathrm{SO}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_{2(\mathrm{~g})}\); is \(0.20 \mathrm{~mole}^{1 / 2}\) litre \(^{-1 / 2}\) at \(1000 \mathrm{~K}\). Calculate equilibrium constant for $$ 2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}=2 \mathrm{SO}_{3(\mathrm{~g})} $$

Chemical equilibrium is a dynamic state in which the rates of the forward and reverse reactions are equal, causing no net change in the concentrations of reactants and products over time. This concept is vital in understanding how chemical reactions reach a state of balance and how they can be disturbed or maintained.

For instance, the equilibrium state in the reaction \( \mathrm{SO}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{2(\mathrm{~g})} + \frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \) does not imply that all reactants and products cease to change; instead, they do so at a rate that does not affect the overall concentration. When a system is at equilibrium, it can be described quantitatively by an equilibrium constant \( K \) which offers insight into the position of equilibrium and the proportions of reactants and products.

Reaction stoichiometry involves the quantitative relationship between the amounts of reactants and products in a chemical reaction. It is based on the law of conservation of mass, where the total mass of reactants must equal the total mass of products.

Understanding stoichiometry is essential when determining how different equations relate to each other. For instance, comparing the stoichiometry of the given reaction \( \mathrm{SO}_{3} \leftrightarrow \mathrm{SO}_{2} + \frac{1}{2} \mathrm{O}_{2} \) with the target reaction \( 2 \mathrm{SO}_{2} + \mathrm{O}_{2} = 2 \mathrm{SO}_{3} \) shows that the target reaction involves twice the amount of each substance. Consequently, the stoichiometric coefficients directly affect how we calculate the equilibrium constant for the reaction. To adjust the equilibrium constant for a modified reaction stoichiometry, we raise the constant to the power of the coefficient used in the balanced equation, in this case, squaring it because the reaction is reversed and the quantities of products and reactants are doubled.

The reciprocal of the equilibrium constant represents the equilibrium constant for the reverse chemical reaction. When we have an equilibrium constant for a certain reaction, and we need to find the equilibrium constant for the reaction that proceeds in the opposite direction, we simply calculate the reciprocal of the original constant.

In the provided exercise, the equilibrium constant of the reverse reaction was needed to determine the equilibrium constant for the target reaction. The mathematical operation is straightforward: take the inverse of the provided equilibrium constant \( K_{\text{given}} \) to get \( K_{\text{reverse}} = \frac{1}{K_{\text{given}}} \). This step is critical and must be precisely understood and applied when dealing with reactions moving in the opposite direction to those for which the equilibrium constants are given, as in the case of our original equilibrium reaction \( \mathrm{SO}_{3} \leftrightarrow \mathrm{SO}_{2} + \frac{1}{2} \mathrm{O}_{2} \) when analyzing the formation of \( \mathrm{SO}_{3} \) from \( \mathrm{SO}_{2} \) and \( \mathrm{O}_{2} \).