Question

For the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ it is found that at equilibrium \(\left[\mathrm{CO}_{2}\right]=2.1 \times 10^{-3} \mathrm{M}\) at a particular temperature. Calculate \(K\) for the reaction at this temperature.

Short answer

The equilibrium constant, K, for the reaction \(\mathrm{CaCO_{3}(s) \rightleftharpoons CaO(s)+CO_{2}(g)}\) at this temperature is \(K = 2.1 \times 10^{-3}\).

Step-by-step solution

Write the expression for K

Since we only have the concentration of CO₂ to consider, the expression for K will be simplified as follows: \[ K = [\mathrm{CO}_{2}] \]

Plug in the given concentration of CO₂

We are given that the concentration of CO₂ at equilibrium is: \(\ [\mathrm{CO}_{2}] = 2.1 \times 10^{-3} \mathrm{M} \) Now, substitute this value into the K expression: \[ K = 2.1 \times 10^{-3} \]

Find the value of K

Since there is no need for further calculations, the equilibrium constant, K, for this reaction at this temperature is: \[ K = 2.1 \times 10^{-3} \] Thus, the equilibrium constant for the given reaction at this temperature is \(2.1 \times 10^{-3}\).

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when a reversible reaction proceeds in such a way that the rate of the forward reaction equals the rate of the reverse reaction, leading to no overall change in the concentrations of reactants and products over time.

It's essential to note that equilibrium does not mean the reactants and products are in equal concentrations, but rather that their concentrations have stabilized in a ratio that will not change without external influence. In the example given, with the decomposition of \(\mathrm{CaCO}_{3}\) to form \(\mathrm{CaO}\) and \(\mathrm{CO}_{2}\), the system reaches a point where, for every molecule of \(\mathrm{CaCO}_{3}\) that decomposes, there's a molecule of \(\mathrm{CaO}\) and \(\mathrm{CO}_{2}\) that recombine back to form \(\mathrm{CaCO}_{3}\).

Moreover, at equilibrium, the concentration of products and reactants remains constant over time. In our equation, since \(\mathrm{CaCO}_{3}\) and \(\mathrm{CaO}\) are solids, their concentrations do not appear in the equilibrium expression, resulting in a simplified expression where only the concentration of \(\mathrm{CO}_{2}\), a gas, is taken into account. Therefore, understanding how to find the equilibrium constant, \(K\), helps us quantify the position of equilibrium at various conditions.

Calculating the Reaction Quotient

The reaction quotient, \(Q\), is an essential tool used to predict the direction a reaction will move to reach equilibrium. It is calculated in the same way as the equilibrium constant, \(K\), using the concentrations of the reactants and products at any point in time before the system has reached equilibrium.

However, unlike \(K\), which remains constant for a given reaction at a specific temperature, the value of \(Q\) changes as the reaction progresses. If \(QK\), the reverse reaction is favored, and reactants will form until equilibrium is reached. When \(Q=K\), the reaction is at equilibrium, and no further net change will occur.

In the exercise, we only refer to \(K\) because we are dealing with conditions at equilibrium, but understanding \(Q\) is also beneficial to predict how the system will respond when not at equilibrium and how it will shift to achieve that equilibrium.

Applying Le Chatelier's Principle

Le Chatelier's principle is named after the French chemist Henri Louis Le Chatelier and it provides a qualitative way of understanding how a change in conditions can affect the position of equilibrium. According to this principle, if an external change is imposed on a system at equilibrium, the system will adjust itself to counteract that change and a new equilibrium position will be established.

Some common changes that can affect the equilibrium position include changes in concentration, pressure, and temperature. For instance, if the pressure is increased by decreasing the volume of a gaseous system, the equilibrium will shift in the direction that produces fewer moles of gas. Similarly, adding more of a reactant or removing a product will shift the equilibrium to produce more products, as seen in our reaction with \(\mathrm{CaCO}_{3}\).

Le Chatelier's principle is particularly useful for understanding how to manipulate conditions to favor the formation of a desired product in chemical processes. By applying this principle, chemists can optimize conditions to increase the yield of reactions efficiently.