Question

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\). b. \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\). c. \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\). d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\).

Short answer

a. \(K = \frac{[\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{NH}_{3}]^{2} [\mathrm{CO}_{2}}\), \(K_{p} = \frac{P_{\mathrm{H}_{2}\mathrm{O}}}{P_{\mathrm{NH}_{3}}^{2} P_{\mathrm{CO}_{2}}}\) b. \(K = [\mathrm{N}_{2}] [\mathrm{Br}_{2}]^{3}\), \(K_{p} = P_{\mathrm{N}_{2}} P_{\mathrm{Br}_{2}}^{3}\) c. \(K = [\mathrm{O}_{2}]^{3}\), \(K_{p} = P_{\mathrm{O}_{2}}^{3}\) d. \(K = \frac{[\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{H}_{2}]}\), \(K_{p} = \frac{P_{\mathrm{H}_{2}\mathrm{O}}}{P_{\mathrm{H}_{2}}}\)

Step-by-step solution

a. Write the expression for K and Kp for 2NH3(g) + CO2(g) ⇌ N2CH4O(s) + H2O(g)

First, we need to identify the gaseous substances in this reaction. They are NH3, CO2, and H2O. N2CH4O is a solid and will not be included in the expressions for K and Kp. The expression for K is given by: \(K = \frac{[\mathrm{H}_{2}\mathrm{O}]^{1}}{[\mathrm{NH}_{3}]^{2} [\mathrm{CO}_{2}]^{1}}\) The expression for Kp is given by: \(K_{p} = \frac{P_{\mathrm{H}_{2}\mathrm{O}}^{1}}{P_{\mathrm{NH}_{3}}^{2} P_{\mathrm{CO}_{2}}^{1}}\)

b. Write the expression for K and Kp for 2NBr3(s) ⇌ N2(g) + 3Br2(g)

We notice that the only gaseous substances in this reaction are N2 and Br2. NBr3 is a solid and will not be included in the expressions for K and Kp. The expression for K is given by: \(K = \frac{[\mathrm{N}_{2}]^{1} [\mathrm{Br}_{2}]^{3}}{1}\) The expression for Kp is given by: \(K_{p} = \frac{P_{\mathrm{N}_{2}}^{1} P_{\mathrm{Br}_{2}}^{3}}{1}\)

c. Write the expression for K and Kp for 2KClO3(s) ⇌ 2KCl(s) + 3O2(g)

The only gaseous substance in this reaction is O2. KClO3 and KCl are both solids and will not be included in the expressions for K and Kp. The expression for K is given by: \(K = \frac{[\mathrm{O}_{2}]^{3}}{1}\) The expression for Kp is given by: \(K_{p} = \frac{P_{\mathrm{O}_{2}}^{3}}{1}\)

d. Write the expression for K and Kp for CuO(s) + H2(g) ⇌ Cu(l) + H2O(g)

In this reaction, the gaseous substances are H2 and H2O. CuO is a solid and Cu is a liquid, and neither will be included in the expressions for K and Kp. The expression for K is given by: \(K = \frac{[\mathrm{H}_{2}\mathrm{O}]^{1}}{[\mathrm{H}_{2}]^{1}}\) The expression for Kp is given by: \(K_{p} = \frac{P_{\mathrm{H}_{2}\mathrm{O}}^{1}}{P_{\mathrm{H}_{2}}^{1}}\)

Key concepts

Equilibrium Constant (K)

The equilibrium constant, denoted as K, is a crucial concept in chemical equilibrium that signifies the ratio of the concentration of the products to the reactants, each raised to the power of their coefficients in a balanced chemical equation. It's essential to remember that K is determined only by gaseous and aqueous species, while solids and liquids are omitted since their concentrations remain constant under given conditions.

For example, in reaction a from the exercise, K is defined strictly using the concentrations of the gaseous ammonia (NH₃) and carbon dioxide (CO₂), as well as the gaseous product water (H₂O), which are involved in the dynamic equilibrium. Similarly, for reaction b, gaseous nitrogen (N₂) and bromine (Br₂) dictate the value of K. Such expressions are pivotal for predicting the direction and extent of chemical reactions under constant temperature conditions.

Partial Pressure (Kp)

Just like the equilibrium constant K, the equilibrium constant for partial pressures, denoted as K_p, provides a snapshot of a system's position at equilibrium, but specifically for gas-phase reactions. It relates to the partial pressures of the gases involved, rather than their concentrations. The formula for K_p is derived from the ideal gas law, which connects concentration and pressure.

When calculating K_p, as shown in the solutions for reactions a, b, and d, you utilize the partial pressures of the gaseous reactants and products. For instance, in reaction c, since oxygen (O₂) is the only gaseous species produced, the expression for K_p solely depends on the partial pressure of oxygen.

Le Chatelier's Principle

When a dynamic equilibrium is disturbed, Le Chatelier's principle helps us predict how the system will adjust to re-establish equilibrium. This principle implies that if an external stress is applied, like a change in concentration, pressure, or temperature, the equilibrium will shift in a direction that counteracts the change.

External Stress and Equilibrium Shift

For example, if you increase the concentration of reactants in a closed system, the principle suggests that the equilibrium will shift towards the products to reduce the added reactants. Similarly, changes in pressure will affect systems with gaseous reactants and products differently depending on the total number of moles of gas on each side of the balanced equation.

Reactants and Products in Equilibrium

In an equilibrium state, the rate of the forward reaction equals the rate of the reverse reaction, maintaining a constant ratio of reactants to products as represented by the equilibrium constant K. This constant state doesn't imply that the reactions have stopped; rather, they occur continuously, but at equivalent rates, thus maintaining stable concentrations of reactants and products.

Understanding the concept of equilibrium helps in manipulating conditions to favor the formation of products or reactants, as desired in industrial processes. It's also noteworthy that equilibrium constants are a valuable tool in equilibrium calculations, allowing chemists to predict concentrations of species in a reaction mixture at equilibrium.