Question

1 mole of \(\mathrm{NO}\) and 1 mole of \(\mathrm{O}_{3}\) are taken in a \(10 \mathrm{~L}\) vessel and heated. At equilibrium, \(50 \%\) of \(\mathrm{NO}\) (by mass) reacts with \(\mathrm{O}_{3}\) according to the equation: $$ \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} $$ What will be the equilibrium constant for this reaction? (a) 1 (b) 2 (c) 3 (d) 4

Short answer

The equilibrium constant (Kc) for this reaction is 1.

Step-by-step solution

Determine the initial moles

Initially, there is 1 mole of NO and 1 mole of O3 each in a 10 L vessel. Thus, the initial concentration of NO and O3 is 0.1 M (1 mole/10 L) for each.

Calculate the change in concentration

At equilibrium, 50% of NO reacts. Since we start with 1 mole of NO, 0.5 mole of NO is left unreacted at equilibrium. Thus, 0.5 mole of NO, O3, NO2, and O2 are present at equilibrium since the stoichiometry of the reaction is 1:1.

Write the equilibrium expression

The equilibrium constant expression for the reaction is given by Kc = [NO2][O2] / ([NO][O3]). At equilibrium, each of these species has a concentration of 0.05 M (0.5 mole/10 L).

Plug concentrations into the equilibrium expression

Substitute the equilibrium concentrations into the expression: Kc = (0.05 * 0.05) / (0.05 * 0.05) = 1.

Conclude the equilibrium constant

The equilibrium constant, Kc, for this reaction is 1, which corresponds to option (a).

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs in a chemical reaction when 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, even though the reactions continue to occur. In the given exercise, the mixture of NO and O3 in a 10 L vessel eventually reaches a state where 50% of NO has reacted with O3 to form NO2 and O2.

It's crucial to understand that at equilibrium, the quantity of NO and O3 that have reacted is equal to the quantity of NO2 and O2 formed, due to the stoichiometric ratio of 1:1 in the reaction. It's also important to recognize that equilibrium does not mean that the reactants and products are in equal concentrations, but that their concentrations no longer change with time. For the example provided, chemical equilibrium is achieved when there is 0.5 mole of NO, O3, NO2 and O2 each in the container.

Reaction Quotient

The reaction quotient, Q, is a measure that tells us whether a reaction at a given moment is at equilibrium, and if not, in which direction it needs to shift to reach equilibrium. It is calculated in the same way as the equilibrium constant, Kc, using the concentrations of the reactants and products at any point in time. However, while Kc is specific to a particular temperature and describes the system at equilibrium, Q can be calculated at any stage of the reaction.

The formula for Q is similar to Kc, involving the concentrations of the products raised to the power of their stoichiometric coefficients divided by the concentrations of the reactants raised to the power of their stoichiometric coefficients. If Q is less than Kc, the reaction will proceed forward to reach equilibrium. If Q is greater than Kc, the reaction will proceed in the reverse direction. And if Q equals Kc, the system is at equilibrium, as illustrated in the exercise's solved example.

Le Chatelier's Principle

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle helps us predict the effect of changing conditions — such as concentration, pressure, and temperature — on the position of equilibrium in a chemical reaction.

For a reactant or product in a gaseous state, an increase in pressure, by decreasing volume, will shift the equilibrium position towards the side with fewer moles of gas. Conversely, a decrease in pressure, by increasing volume, will favor the direction with more moles of gas. In the context of the exercise, if additional NO or O3 were added to the system, the equilibrium would shift to produce more NO2 and O2 to return to equilibrium. If NO2 or O2 were removed, the equilibrium would shift to produce more of these products from the remaining NO and O3.