Question

Consider the reaction $$2 \operatorname{CO}(g)+\mathrm{O}_{2}(g)=2 \operatorname{CO}_{2}(g)$$ Suppose the system is already at equilibrium, and then an additional mole of \(\mathrm{CO}(g)\) is injected into the system at constant temperature. Does the amount of \(\mathrm{CO}_{2}(g)\) in the system increase or decrease? Does the value of \(K\) for the reaction change?

Short answer

Upon adding an extra mole of CO(g) to the system at equilibrium, the reaction shifts in the forward direction to counteract the disturbance. As a result, the amount of CO2(g) in the system increases. However, the value of the equilibrium constant K remains unchanged as the temperature is constant.

Step-by-step solution

Identify the reaction and equilibrium constant

The given reaction is: \[2 \operatorname{CO}(g)+\mathrm{O}_{2}(g)=2 \operatorname{CO}_{2}(g)\] We will use the reaction quotient Q and the equilibrium constant K to describe the system.

Understand Le Chatelier's Principle

According to Le Chatelier's Principle, when a system at equilibrium is subjected to a stress, the system will shift to counteract the stress and minimize the disturbance. In this case, the stress is the addition of extra CO(g) to the system. To counteract this, the system will shift either in the forward or backward direction.

Determine the direction of the shift

As the system is at equilibrium and we have added extra CO(g), the concentration of CO(g) increases. To counteract this, the reaction will shift in the forward direction, which means the system will consume more CO(g) and O2(g) to produce more CO2(g).

Determine the change in the amount of CO2(g)

Since the reaction has shifted forward, it means that more CO2(g) will be produced. Therefore, the amount of CO2(g) in the system will increase.

Determine the change in the value of K

The value of the equilibrium constant K depends only on temperature. Since the temperature of the system remains constant, the value of K will not change. To summarize, the addition of one mole of CO(g) to the system at equilibrium will cause an increase in the amount of CO2(g) due to the shift in the forward direction to minimize the disturbance. However, the value of K remains unchanged since the temperature is constant.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a reliable guide for predicting the effect of changes in concentration, temperature, and pressure on a chemical system at equilibrium. In simple terms, this principle states that if an external change is imposed on a system at equilibrium, the system will react in such a way to counteract the change.

Let's say, for instance, we have a sealed bottle containing a mixture of nitrogen and hydrogen gases in a dynamic equilibrium with ammonia. If we were to add more nitrogen into the bottle, according to Le Chatelier's Principle, the system would respond by shifting the equilibrium to reduce the concentration of nitrogen. In other words, more ammonia would be produced as the system seeks a new balance point.

In the context of the reaction \(2 \text{CO}(g) + \text{O}_2(g) = 2 \text{CO}_2(g)\), injecting additional \(\text{CO}(g)\) into the system at equilibrium provokes a shift to the right, leading to an increased production of \(\text{CO}_2(g)\). This concept is crucial for anyone studying chemistry, as it provides insight into how changes in conditions affect chemical reactions.

Equilibrium Constant (K)

The equilibrium constant, designated as K, is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction at a constant temperature. The formula is written with the concentrations of the products in the numerator and the concentrations of the reactants in the denominator, each raised to the power of their stoichiometric coefficients.

In a general sense, if K is a large number, it indicates that the products are favored at equilibrium, while a small K value suggests that the reactants are favored. For the previous example, where the reaction is \(2 \text{CO}(g) + \text{O}_2(g) = 2 \text{CO}_2(g)\), the equilibrium constant would take the form \(K = \frac{[\text{CO}_2]^2}{[\text{CO}]^2[\text{O}_2]}\).

To clarify, the value of K is not affected by changes in concentration of reactants or products as it solely depends on temperature. This is important because it means even though the concentrations of \(\text{CO}\), \(\text{O}_2\), and \(\text{CO}_2\) shift to re-establish equilibrium after a disturbance, K itself remains the same unless the temperature changes.

Reaction Quotient (Q)

The reaction quotient, Q, serves as a 'snapshot' of the relative concentrations of reactants and products at any given moment in time, not just at equilibrium. It uses the same formula as the equilibrium constant, K, but applies to any concentrations, not just those at equilibrium. By comparing the values of Q and K, we can predict the direction that a reaction will proceed to reach equilibrium.

If \(Q < K\), the reaction will proceed in the forward direction to produce more products. Conversely, when \(Q > K\), there is an excess of products, and the reaction will shift in the reverse direction, forming more reactants. If \(Q = K\), the system is at equilibrium and no further shift will occur.

When the additional mole of \(\text{CO}(g)\) is injected into the equilibrium system in our original exercise, Q suddenly becomes less than K, because the ratio of the reactant concentration has increased. Consequently, the reaction is driven to the right, creating more \(\text{CO}_2(g)\) as the system strives to restore equilibrium.