Problem 3 - Chemistry An Atoms First Approach

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.

Step-by-Step Solution

Answer

The final equilibrium mixtures will be different for the two procedures as they have different initial conditions. However, the equilibrium constant will remain the same for both cases, as the system adjusts the concentration of the species to reach equilibrium according to Le Châtelier's principle.

Step 1: Understand the Reaction

We are given the following reaction: \[ \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \] In both cases, we need to figure out the equilibrium concentrations of each species, and then compare their final equilibrium concentrations. We will follow the procedures (a) and (b), analyze the changes in moles/reactant ratios, and observe the final equilibrium mixtures.

Step 2: Procedure (a) - Initial Equilibrium

For Procedure (a): 1. Initially, we have 0.5 moles of H2 and 0.5 moles of I2. 2. In order to reach equilibrium, let's assume that 'x' moles of H2 and I2 react to form 2x moles of HI. At equilibrium, we have: \[ \mathrm{H}_{2}: 0.5-x \, \mathrm{mol} \] \[ \mathrm{I}_{2}: 0.5-x \, \mathrm{mol} \] \[ \mathrm{HI}: 2x \, \mathrm{mol} \]

Step 3: Procedure (a) - Adding 1 Mole of H2

Now, we add another mole of H2 and let the system reach equilibrium again. Assume that 'y' moles of H2 and I2 react to form 2y moles of HI in this new equilibrium. At this new equilibrium, we have: \[ \mathrm{H}_{2}: (0.5-x) + 1 - y \, \mathrm{mol} \] \[ \mathrm{I}_{2}: (0.5-x) - y \, \mathrm{mol} \] \[ \mathrm{HI}: 2x + 2y \, \mathrm{mol} \]

Step 4: Procedure (b) - Initial Equilibrium

For Procedure (b): 1. Initially, we have 1.5 moles of H2 and 0.5 moles of I2. 2. In order to reach equilibrium, let's assume that 'z' moles of H2 and I2 react to form 2z moles of HI. At equilibrium, we have: \[ \mathrm{H}_{2}: 1.5-z \, \mathrm{mol} \] \[ \mathrm{I}_{2}: 0.5-z \, \mathrm{mol} \] \[ \mathrm{HI}: 2z \, \mathrm{mol} \]

Step 5: Comparing Final Equilibrium Mixtures

Now let's compare the final equilibrium mixtures from both the procedures: Procedure (a): \[ \mathrm{H}_{2}: (0.5-x) + 1 - y \, \mathrm{mol} \] \[ \mathrm{I}_{2}: (0.5-x) - y \, \mathrm{mol} \] \[ \mathrm{HI}: 2x + 2y \, \mathrm{mol} \] Procedure (b): \[ \mathrm{H}_{2}: 1.5-z \, \mathrm{mol} \] \[ \mathrm{I}_{2}: 0.5-z \, \mathrm{mol} \] \[ \mathrm{HI}: 2z \, \mathrm{mol} \] Both the cases lead to different equilibrium mixtures. However, the equilibrium constant remains the same under the same experimental conditions. The reaction quotient (Q) will initially be different in both procedures. As the system adjusts the concentration of the species to reach equilibrium, the final Q value will equal the equilibrium constant (K), according to Le Châtelier's principle. In conclusion, the final equilibrium mixtures will indeed be different for the two procedures as they have different initial conditions. However, the equilibrium constant will remain the same for both cases.

Key Concepts

Le Châtelier's Principle

Le Châtelier's Principle is an essential concept in understanding how chemical equilibria respond to changes in conditions. According to this principle, if a dynamic equilibrium system experiences a disturbance, such as a change in concentration, temperature, or pressure, the system will adjust to counteract the disturbance and restore a new equilibrium. This adjustment involves shifting the balance between the reactants and products.

In the given exercise, when 1 mole of \( \mathrm{H}_2 \) is added to the equilibrium mixture in procedure (a), the system undergoes a disturbance. Le Châtelier's Principle predicts that the reaction will shift to favor the forward reaction, converting more \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \) into \( \mathrm{HI} \), to partially counteract the change caused by adding \( \mathrm{H}_2 \).

The final equilibrium achieved will, therefore, be altered as the amounts of substances in the system are different compared to when no \( \mathrm{H}_2 \) was added. Hence, this principle provides a qualitative prediction of how the addition of reactants or products affects the equilibrium composition.

Equilibrium Constant

The equilibrium constant \( K \) is a numerical value that characterizes the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It is expressed in terms of the concentration of each species involved in the reaction. For the given reaction \( \mathrm{H}_2 + \mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI} \), the equilibrium constant expression is given by:

\[ K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \]

The value of \( K \) is constant for a specific reaction at a given temperature, regardless of the initial starting concentrations. This implies that no matter the route taken to reach equilibrium (as in procedures (a) and (b)), \( K \) remains unchanged. In practice, this means that while the equilibrium concentrations might differ due to different starting conditions, the ratio governed by \( K \) stays the same, ensuring consistency in the system's equilibrium behavior.

Reaction Quotient

The reaction quotient \( Q \) is similar in form to the equilibrium constant \( K \), but it applies to any set of concentrations, not just at equilibrium. It helps predict the direction in which a reaction will shift to reach equilibrium.

Calculation of \( Q \) is done using the current concentrations of the reactants and products just like the equilibrium constant formula:

\[ Q = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \]

If \( Q = K \), the system is at equilibrium.

If \( Q < K \), the reaction will proceed forward, resulting in more products being formed.

If \( Q > K \), the reaction will shift backward, favoring the formation of reactants.

In procedures (a) and (b) from the exercise, the initial values of \( Q \) will be different due to varying starting amounts of \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \). As the system progresses towards equilibrium, \( Q \) will adjust to match \( K \), dictating the direction of the shifts.

Equilibrium Calculations

Equilibrium calculations are essential for quantifying the concentrations of reactants and products at equilibrium. They often involve setting up an 'ICE' table, which stands for Initial, Change, and Equilibrium. This method provides a structured approach to solving equilibrium problems.

Initially, list the starting concentrations (or moles) of each reactant and product. Then, define the changes in concentration that occur as the system moves towards equilibrium. Finally, express the equilibrium concentrations using these changes. In mathematical terms, you can represent this change through the variable \( x \), representing the amount of reactants converted into products at equilibrium.

In procedure (a), for example, the initial step involves solving for 'x' to determine the equilibrium concentrations of \( \mathrm{H}_2 \), \( \mathrm{I}_2 \), and \( \mathrm{HI} \). Using the relation \( 0.5 - x \) for reactants and \( 2x \) for \( \mathrm{HI} \), you can find these values. When 1 mole of \( \mathrm{H}_2 \) is added, a new set of equilibrium calculations is required to understand how the system readjusts.

Performing these calculations accurately allows us to compare outcomes for different procedures in the exercise and understand the inherent nature of chemical equilibria.