Problem 65 - Chemistry

Question

Determine the initial and equilibrium concentrations of HI if the initial concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) are both \(0.16 M\) and their equilibrium concentrations are both \(0.072 M\) at \(430^{\circ} \mathrm{C}\). The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) is 54.2 at \(430^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The initial concentration of HI is 0 M and the equilibrium concentration of HI is 0.176 M.

Step 1: Analyze the Reaction and Write the Equilibrium Expression

The reaction given is \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\). The equilibrium constant expression for this reaction is \(K_{c} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}\). This is derived from the balanced chemical equation. Because the coefficient of HI is 2, its concentration is squared in the equilibrium equation.

Step 2: Calculate the Change in Concentration

Both the initial concentrations of H2 and I2 decrease by 0.088 M, therefore, the concentration of HI would increase by 2*0.088 M = 0.176 M because 2 moles of HI are produced from 1 mole each of H2 and I2. Hence, for every 0.088 M used up, 0.176 M is produced.

Step 3: Determine the Equilibrium Concentrations

Adding the change to the starting concentration gives the equilibrium concentration of HI as 0 M + 0.176 M = 0.176 M.

Step 4: Check With the Equilibrium Constant

Substitute all the equilibrium concentrations into the equilibrium expression obtained in step 1 and solve for the equilibrium constant. If the calculated value matches the given value of Kc, then the equilibrium concentration is correct. Here, \(K_{c} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}= \dfrac{(0.176)^2 }{(0.072) \times (0.072)} = 54.2\). As these values match exactly, it can be confirmed that the equilibrium concentrations are correct!

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, often represented as \(K_c\), is a crucial concept. It provides a measure of how far a reaction will proceed at a given temperature. For reversible reactions, like the one between hydrogen and iodine forming hydrogen iodide, some amount of reactants remain even after the reaction has reached equilibrium. The equilibrium constant is calculated using the concentrations of the reactants and products at equilibrium. In our case, \(K_c = 54.2\) implies that, at \(430^{\circ}\)C, the products are favored significantly. High \(K_c\) values typically indicate that the formation of products is favored when the system has reached equilibrium. This doesn't mean the reaction goes to completion, but rather that products dominate over reactants at the equilibrium point.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. It can proceed in either a forward or backward direction, especially when it is a reversible reaction, as seen in our example with hydrogen gas reacting with iodine gas to form hydrogen iodide: \(\mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2\mathrm{HI}(g)\). This equation signifies the process where:

• One molecule of \(\mathrm{H}_2\) reacts with one molecule of \(\mathrm{I}_2\).
• Two molecules of \(\mathrm{HI}\) are formed.
• The double arrow indicates that the reaction can proceed in both directions, reaching equilibrium when the rate of forward reaction equals the rate of the reverse reaction.

This balance is key to understanding how equilibrium concentrations are determined, ensuring that despite changes in concentration, the reaction maintains its equilibrium state.

Initial Concentration

The initial concentration is the starting point for determining how a reaction progresses towards equilibrium. For our reaction, the initial concentrations of \(\mathrm{H}_2\) and \(\mathrm{I}_2\) were both \(0.16 M\). These initial values are important because they help establish how much of each reactant will be consumed and how much product will be formed as the system reaches equilibrium. The difference between the initial and equilibrium concentrations provides insight into the reaction's progress. Initial concentrations set the stage for calculating changes in concentration, driving the reaction towards its equilibrium state.

Change in Concentration

Change in concentration refers to how the amounts of reactants and products alter as the reaction moves from its initial state to equilibrium. In our example, the concentration of \(\mathrm{H}_2\) and \(\mathrm{I}_2\) decreases from \(0.16 M\) to \(0.072 M\). This decrease is recorded as \(0.088 M\) for both reactants, indicating how much was consumed to form the product. Consequently, the concentration of \(\mathrm{HI}\) increases by \(2 \times 0.088 M = 0.176 M\).

• This demonstrates the stoichiometry of the reaction: two moles of \(\mathrm{HI}\) are produced for each mole of \(\mathrm{H}_2\) and \(\mathrm{I}_2\) consumed.
• Understanding changes in concentration is essential for calculating equilibrium concentrations and confirming the consistency with equilibrium constants.

These changes are the bridge between initial conditions and equilibrium states, showcasing the dynamism inherent in chemical reactions.