Calculating Equilibrium Concentrations For H₂ I₂ And HI Reaction At 300°C

In the realm of chemical kinetics and thermodynamics, the concept of chemical equilibrium holds paramount importance. It dictates the extent to which a reversible reaction proceeds and the relative amounts of reactants and products present at the state of equilibrium. Let's delve into a fascinating example, the reversible reaction between hydrogen (H₂) and iodine (I₂) to form hydrogen iodide (HI), under specific conditions, and analyze the equilibrium composition.

The Reaction Scenario: H₂(g) + I₂(g) ⇌ 2 HI(g)

Consider a closed 1-liter container, initially at room temperature, housing 1 mole of iodine (I₂) and 1 mole of hydrogen (H₂). The system is then heated to 300°C, a temperature at which the reaction between H₂ and I₂ to form HI proceeds at a measurable rate. This reaction is reversible, meaning that HI can also decompose back into H₂ and I₂. The equilibrium constant, denoted as Kc, for this reaction at 300°C is given as 1.0 x 10². This high value indicates that at equilibrium, the reaction mixture will predominantly consist of the product, HI.

To truly understand this equilibrium, it's crucial to consider the factors influencing it. Temperature plays a pivotal role, as chemical equilibrium is temperature-dependent. The given temperature of 300°C is not merely a detail; it's a critical parameter that governs the balance between reactants and products. The equilibrium constant (Kc) itself is a function of temperature, highlighting the dynamic interplay between thermodynamics and reaction kinetics. The high value of Kc at this temperature suggests that the formation of HI is highly favored, but it doesn't guarantee complete conversion. The system will strive to reach a state where the rates of the forward and reverse reactions are equal, leading to a stable mixture of H₂, I₂, and HI. This dynamic state is what we refer to as chemical equilibrium, a state where macroscopic properties remain constant, but microscopic activity persists.

Calculating Equilibrium Concentrations

To quantify the composition at equilibrium, let's define 'x' as the amount (in moles) of H₂ and I₂ that react to form HI. Using the stoichiometry of the reaction, we can deduce the changes in the amounts of each species:

• Initial moles:
  • [H₂] = 1 mol
  • [I₂] = 1 mol
  • [HI] = 0 mol
• Change in moles:
  • [H₂] = -x mol
  • [I₂] = -x mol
  • [HI] = +2x mol
• Equilibrium moles:
  • [H₂] = (1 - x) mol
  • [I₂] = (1 - x) mol
  • [HI] = 2x mol

Since the volume of the container is 1 liter, the concentrations are numerically equal to the number of moles. The equilibrium constant, Kc, is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients:

Kc = [HI]² / ([H₂] [I₂])

Substituting the equilibrium concentrations into the Kc expression, we get:

1. 0 x 10² = (2x)² / ((1 - x) (1 - x))

This equation allows us to solve for 'x', which represents the extent of the reaction at equilibrium. The mathematical solution of this quadratic equation provides a quantitative measure of how much hydrogen and iodine have reacted to form hydrogen iodide.

Solving for 'x' and Equilibrium Composition

The equation derived from the equilibrium constant expression is a quadratic equation, and solving it will provide the value of 'x'. However, given the high value of Kc (1.0 x 10²), we can make a simplifying assumption to ease the calculation. Since Kc is large, it suggests that the reaction proceeds almost to completion. This implies that 'x' will be close to 1, meaning most of the H₂ and I₂ will react to form HI. Therefore, we can assume that (1 - x) is a small value.

However, to avoid inaccuracies, it's best to solve the quadratic equation directly. Expanding the equation:

100 = (4x²) / (1 - 2x + x²)

100 - 200x + 100x² = 4x²

96x² - 200x + 100 = 0

This quadratic equation can be solved using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Where a = 96, b = -200, and c = 100.

Solving for x gives us two possible values, but only one will be physically meaningful (between 0 and 1, as x represents the fraction of H₂ and I₂ that react). The value of x is approximately 0.926.

Now, we can calculate the equilibrium concentrations:

• [H₂] = 1 - x = 1 - 0.926 ≈ 0.074 M
• [I₂] = 1 - x = 1 - 0.926 ≈ 0.074 M
• [HI] = 2x = 2 * 0.926 ≈ 1.852 M

These equilibrium concentrations reveal the composition of the mixture at 300°C. The concentration of HI is significantly higher than that of H₂ and I₂, reflecting the large equilibrium constant and the favorable formation of HI at this temperature.

Interpretation of Results

The equilibrium concentrations provide a quantitative picture of the system at equilibrium. The high concentration of HI compared to H₂ and I₂ confirms our earlier prediction based on the large Kc value. This indicates that the reaction strongly favors the formation of the product, HI, under these conditions.

The relatively small but non-zero concentrations of H₂ and I₂ are also significant. They highlight the dynamic nature of equilibrium. Even though the reaction favors HI formation, the reverse reaction (decomposition of HI) still occurs to some extent, maintaining a balance between reactants and products. The equilibrium is not a static state but rather a dynamic balance where the rates of the forward and reverse reactions are equal.

Factors Affecting Equilibrium: Le Chatelier's Principle

This analysis provides a snapshot of the equilibrium state at 300°C. However, it's essential to recognize that equilibrium is sensitive to changes in conditions. Le Chatelier's principle provides a qualitative framework for predicting how changes in conditions will affect equilibrium.

Le Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. The