Problem 3 - Chemistry

Question

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),\) consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or \((b)\) you \(\operatorname{mix} 1.5\) moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

Step-by-Step Solution

Answer

The final equilibrium mixtures for both procedures will be the same. This is because the equilibrium state depends on the initial concentrations of reactants/products and the equilibrium constant. According to Le Chatelier's principle, the system will adjust its rates of reaction in order to counteract any disturbances and achieve a new equilibrium state, ultimately resulting in the same final equilibrium mixtures for both procedures.

Step 1: Procedure (a) - Initial Equilibrium

First, let's consider the initial equilibrium system where 0.5 moles of H2 and 0.5 moles of I2 are mixed. Let x be the moles of each reactant consumed to reach the initial equilibrium state, and 2x be the moles of HI formed at the equilibrium state. The final moles at the initial equilibrium will be: H2: 0.5 - x I2: 0.5 - x HI: 2x

Step 2: Procedure (a) - Addition of 1 mole H2

When an additional mole (1.0 moles) of H2 is added, the moles at the start of the second step are: H2: 1.5 - x I2: 0.5 - x HI: 2x

Step 3: Procedure (a) - Second Equilibrium

Let y be the moles of each reactant consumed in the second equilibrium state, and 2y be the moles of HI formed. The final moles at the second equilibrium will be: H2: (1.5 - x) - y I2: (0.5 - x) - y HI: 2x + 2y

Step 4: Procedure (b) - Equilibrium

Now let's consider the other method where 1.5 moles of H2 and 0.5 moles of I2 are mixed directly without adding H2 in steps. Let z be the moles of each reactant consumed to reach the equilibrium state and 2z be the moles of HI formed at the equilibrium state. The final moles at equilibrium will be: H2: 1.5 - z I2: 0.5 - z HI: 2z

Step 5: Comparing Equilibrium Mixtures

Now we need to compare the equilibrium mixtures between these two methods. According to Le Chatelier's principle, when a system at equilibrium is disturbed by a change in concentration, the system will adjust its rates of reaction in order to counteract the disturbance, and regain a new equilibrium state. In Procedure (a), we have disturbed the equilibrium state of the reaction by adding an extra mole of H2. The second equilibrium state will again follow the same equilibrium constant. Since the final equilibrium state depends on the concentration of reactants/products and the equilibrium constant, it will be the same as that of Procedure (b). In other words, both procedures described above will have the same final equilibrium state, as the equilibrium depends on the initial concentrations of reactants/products and the equilibrium constant. Hence, the final equilibrium mixtures for both procedures will be the same.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemical equilibrium. It explains how a system at equilibrium reacts to external changes. The principle states:

• When a change in concentration, pressure, or temperature is applied to a system at equilibrium, the system will adjust to counteract the change and restore a new equilibrium state.

For example, in the context of the exercise, when we add more hydrogen (\(H_2\)) to the reaction system, the principle predicts that the system will shift towards forming more \(HI\). Why does this happen? The system is pushing back against the added \(H_2\) to decrease its concentration.
This shift continues until the reaction again reaches equilibrium, maintaining the equilibrium constant. By understanding this principle, we see that despite disturbances to the system, the eventual equilibrium phenomena remain consistent.

Equilibrium Constant

The equilibrium constant is crucial for understanding the balance of chemical reactions at equilibrium. Denoted as \(K_c\), it relates to the concentrations of reactants and products at equilibrium.

• The expression for \(K_c\) is derived from the balanced chemical equation.
• For the reaction \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g)\), it is given by:\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

The value of \(K_c\) provides insights into the reaction dynamics:

• If \(K_c\) is greater than 1, products are favored, while a \(K_c\) less than 1 indicates a reactant-favored system.

Importantly, \(K_c\) remains unchanged unless temperature changes. Thus, whether we follow procedure (a) or (b) from the original exercise, the \(K_c\) and thereby the eventual equilibrium concentrations, remain consistent.

Reaction Stoichiometry

Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It serves as a recipe for predicting the amounts consumed and created during the reaction.
For the given reaction \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g)\), the stoichiometric coefficients are essential:

• 1 mole of \(H_2\) reacts with 1 mole of \(I_2\) to produce 2 moles of \(HI\).

This relationship helps in setting up the equations to calculate equilibrium positions.
In both procedures outlined in the problem, stoichiometry dictates how changes in initial reactant quantities affect the system but do not alter the equilibrium constant. The stoichiometry tells us exactly how much of each substance changes, but Le Chatelier's Principle and the equilibrium constant determine the system's response.
Thus, understanding stoichiometry alongside these principles allows a more comprehensive prediction of the system's behavior during the equilibrium perturbations.