91Ó°ÊÓ

The equilibrium constant, denoted as \( K \), is a crucial part of understanding chemical equilibrium. It provides insight into the ratio of the concentrations of products to reactants once a chemical reaction has reached equilibrium. In simple terms, it helps us predict the position of equilibrium. If \( K \) is large, it means the reaction heavily favors the products, whereas a small \( K \) indicates the reactants are favored. This particular reaction has \( K = 3.75 \), signifying a moderate preference for the products at equilibrium.
This constant is obtained by inserting the equilibrium concentrations of all substances into the mass action expression. It's important to note that the value of \( K \) is specific to a reaction at a given temperature, and does not change unless the temperature does. Understanding \( K \) allows chemists to estimate how a reaction mixture will behave as conditions change.

As a chemical reaction proceeds towards equilibrium, the concentrations of the reactants and products change over time. In this exercise, starting with equal initial concentrations of 0.800 M for all substances, we observe changes as the system approaches equilibrium. The idea is to define a variable \( x \), which represents the amount by which the concentration of the reactants decreases and that of the products increases.

• Reactants: Their concentrations decrease by \( x \), so for example, the concentration of \( \mathrm{SO}_2 \) changes to \( 0.800 - x \).
• Products: Their concentrations increase by \( x \), making the concentration of \( \mathrm{SO}_3 \) become \( 0.800 + x \).

These calculated changes allow us to substitute the equilibrium concentrations back into the mass action expression to solve for \( x \). This process makes it possible to determine the exact concentrations at equilibrium.

When solving for equilibrium concentrations, we sometimes encounter quadratic equations. This happens particularly when the reaction involves a single-step interaction of one mole of reactants forming one mole of products. After substituting the expressions for concentration changes into the mass action expression, we get a quadratic equation.
In our case, substituting led to the equation:\[ x^2 + 2(0.800)x - (0.800)^2 + \frac{(0.800)^2}{3.75} = 0 \]Quadratics are solved using the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. In chemistry, we often ignore negative roots because concentrations cannot be negative. Thus, solving yields \( x = 0.207 \), which we use to find the concentrations of each component at equilibrium.

The mass action expression is derived from the balanced chemical equation and dictates how the equilibrium constant is expressed mathematically. For a reaction like the one in the exercise, \( \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) \), the expression is set up as:
\[ K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} \].
This ratio shows the relationship between product and reactant concentrations at equilibrium. By substituting the equilibrium concentrations into this expression and solving for the unknown variable \( x \), displacement in concentrations can be calculated. It's a key tool in predicting how different factors such as concentration, pressure, and temperature can affect the equilibrium position.