Equilibrium constants are crucial in chemical reactions, as they help us understand the extent to which a reaction proceeds and the concentrations of reactants and products at equilibrium. In simple terms, the equilibrium constant (K) quantifies the balance between the forward and reverse reactions. Here, we will explore three practical examples to illustrate how equilibrium constants can be applied in real-world scenarios.
The Haber process is a well-known industrial method used to produce ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂). This reaction is crucial for fertilizer production, which supports global agriculture.
In the reaction, the equilibrium constant (K) is expressed as:
K = [NH₃]² / ([N₂][H₂]³)
Where [NH₃], [N₂], and [H₂] are the molar concentrations of ammonia, nitrogen, and hydrogen at equilibrium, respectively.
In a typical industrial setting, the reaction occurs at high pressure and temperature to optimize the yield of ammonia. For example, if the concentrations at equilibrium are [NH₃] = 0.5 M, [N₂] = 0.2 M, and [H₂] = 0.3 M, we can calculate K as follows:
K = (0.5)² / (0.2)(0.3)³ = 2.78
This value indicates that at equilibrium, the production of ammonia is favored under the conditions used in the Haber process.
Acetic acid (CH₃COOH) is a weak acid that partially dissociates in water. Understanding its dissociation constant (Kₐ) is essential for applications in food preservation and various chemical processes.
The equilibrium for the dissociation of acetic acid can be represented as:
CH₃COOH ⇌ H⁺ + CH₃COO⁻
The equilibrium constant is given by:
Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH]
Assuming that at equilibrium, the concentrations are [H⁺] = 0.01 M, [CH₃COO⁻] = 0.01 M, and [CH₃COOH] = 0.98 M, we can calculate Kₐ:
Kₐ = (0.01)(0.01) / (0.98) = 1.02 x 10⁻⁴
This small value indicates that acetic acid is a weak acid, with a low degree of dissociation in solution.
Calcium sulfate (CaSO₄) is a sparingly soluble salt. Understanding its solubility product constant (Kₛₕ) is important in various fields, including construction and medicine.
The equilibrium for the dissolution of calcium sulfate can be represented as:
CaSO₄(s) ⇌ Ca²⁺(aq) + SO₄²⁻(aq)
The solubility product constant is expressed as:
Kₛₕ = [Ca²⁺][SO₄²⁻]
Suppose at equilibrium in a saturated solution, the concentration of Ca²⁺ is 0.01 M and SO₄²⁻ is also 0.01 M. We can then calculate Kₛₕ:
Kₛₕ = (0.01)(0.01) = 1.0 x 10⁻⁴
This indicates that calcium sulfate has low solubility in water, which is significant for its applications in construction materials.