High School

Consider the reaction:

\[ \text{SO}_2(g) + \text{NO}_2(g) \rightarrow \text{SO}_3(g) + \text{NO}(g) \]

The equilibrium constant \( K \) is 18.0 at 110ºC.

If 1.0 mole of \(\text{SO}_2\) and 2.0 moles of \(\text{NO}_2\) are placed in a 20.0 L container, what will be the concentration of \(\text{SO}_3\) at equilibrium?

Answer :

Final answer:

By calculating the initial concentrations of the reactants and setting up the equilibrium constant expression, you can solve for the equilibrium concentration of SO3 using algebra.

Explanation:

The question is asking for the equilibrium concentration of SO3 (sulfur trioxide) in a chemical reaction between SO2 (sulfur dioxide) and NO2 (nitrogen dioxide). In this scenario, we first calculate the initial concentrations of the reactants. The initial concentration of SO2 is 1.0 mole / 20.0 L = 0.05 M, and for NO2 it is 2.0 moles / 20.0 L = 0.10 M. As the reaction progresses towards equilibrium, the concentrations of the reactants decrease and the concentration of the products increase until equilibrium is reached, at which point the concentrations remain constant.

At equilibrium, we can express the concentrations in terms of a change in concentration, represented by 'x'. Based on the stoichiometry of the reaction, if x amount of SO2 reacts, a similar x amount of SO3 is formed. So concentration of SO3 at equilibrium will be 'x'.

The equilibrium constant expression for the reaction is K = [SO3][NO] / [SO2][NO2], and substituting the equilibrium concentrations into this expression gives 18.0 = x(0.10 - x) / (0.05 - x) * (0.10 - x). By solving this equation, you can then find the concentration of SO3 at equilibrium.

Learn more about Equilibrium Concentration here:

https://brainly.com/question/16645766

#SPJ11