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 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.

Short answer

The final equilibrium mixtures for both procedures will not be different. This is because, after the addition of one mole of H₂ in procedure (a), the new concentrations of reactants and products will match with the initial concentrations in procedure (b). Thus, both systems will reach the same final equilibrium regardless of when the additional mole of H₂ is added.

Step-by-step solution

Write the chemical equilibrium equation

The given chemical equilibrium is: \[\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g)\]

Define the Reaction Quotient (Q) and calculate the initial concentrations

Q is defined as: \[Q = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}\] For (a) initially, 0.5 mol of H₂ and 0.5 mol of I₂ are mixed. Suppose the equilibrium constant (K) is given, and an equilibrium concentration of HI is produced. Let x be the decrease in the moles of the reactants and increase in the moles of the products. For (b) initially, 1.5 mol of H₂ and 0.5 mol of I₂ are mixed, and they form an equilibrium concentration of HI.

Calculate the equilibrium concentrations for Procedure (a) and (b) in terms of x

For (a), the equilibrium concentrations are: H₂: (0.5 - x) mol I₂: (0.5 - x) mol HI: 2x mol For (b), the initial concentrations are: H₂: (1.5 - x) mol I₂: (0.5 - x) mol HI: 2x mol

Analyze the changes in the system when more H₂ is added

For (a): when another mole of H₂ is added, the new concentrations are: H₂: [(0.5 - x) + 1] mol = (1.5 - x) mol I₂: (0.5 - x) mol HI: 2x mol Now, compare this to the initial concentrations in (b). They are equal. Therefore, the two systems will reach the same final equilibrium, and there will be no difference in the final equilibrium mixtures. Adding the additional mol of H₂ now in (a) or initially in (b) does not change the final equilibrium position of the reaction.

Key concepts

Le Chatelier's Principle

Understanding Le Chatelier's Principle is crucial to interpret how a chemical reaction responds to changes in conditions. When a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust to partially counteract the effect of that change. This principle can predict the direction in which a reaction will shift to re-establish equilibrium.

For instance, consider a reaction at equilibrium. If the concentration of a reactant is increased, the system will respond by increasing the rate of the reaction that consumes that reactant, thereby producing more products to reduce the reactant's concentration to a new equilibrium state. Le Chatelier's Principle explains why adding an additional mole of H₂ in the given exercise will eventually lead to the same equilibrium state, regardless of whether it is added at the beginning or after the system initially reached equilibrium.

Reaction Quotient (Q)

The Reaction Quotient (Q) plays an instrumental role in determining the direction of the shift to reach chemical equilibrium. Defined as the ratio of the concentrations of the reaction products raised to the power of their respective stoichiometric coefficients to the reactants' concentrations, also raised to their respective stoichiometric coefficients, Q provides a snapshot of a system's status before it reaches equilibrium. The formula for Q is given as:
\[Q = \frac{{[\text{Products}]^\text{coefficients}}}{{[\text{Reactants}]^\text{coefficients}}}\]

In the context of our exercise, Q helps us calculate the initial state of the reaction before additional H₂ is added. By comparing Q to the Equilibrium Constant (K), we can predict whether the reaction will proceed forward, reverse, or stay at equilibrium. For both scenarios (a) and (b), as the system adjusts to changes, Q will evolve until it equals K, signifying that the system has reached equilibrium.

Equilibrium Constant (K)

The Equilibrium Constant (K) is the cornerstone of understanding chemical equilibrium. It is defined for a given reaction at a specific temperature and dictates the proportions of reactants and products at equilibrium. Mathematically, K has the same form as Q but is exclusively calculated at equilibrium:
\[K = \frac{{[\text{Products}]^\text{coefficients}}}{{[\text{Reactants}]^\text{coefficients}}}\]
Where the concentration terms are those at equilibrium. The value of K indicates the extent of the reaction; a larger K favors product formation, while a smaller K favors the reactants.

In the exercise provided, regardless of the initial amounts of H₂ and I₂, the final equilibrium concentrations will adjust to satisfy the constant K value for this reaction. Therefore, the initial conditions in (a) and (b) lead to an equilibrium state characterized by the same ratio of concentrations as dictated by K. Hence, adding more H₂ at the start or after the initial equilibrium will not affect the ratio of products to reactants as long as the external conditions remain constant.