Problem 55 - Chemistry The Molecular Science

Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ has the value 50.0 at \(745 \mathrm{~K}\). (a) When \(1.00 \mathrm{~mol} \mathrm{I}_{2}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) are allowed to come to equilibrium at \(745 \mathrm{~K}\) in a sealed 10.00 -L flask, calculate the amount (in moles) of HI produced. (b) Calculate the amount of HI produced in a 5.00-L flask. (c) Calculate the total amount of HI present at equilibrium if an additional \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is added to the \(10.00-\mathrm{L}\) flask.

Step-by-Step Solution

Answer

(a) Solve for HI in a 10.00-L flask using \( K_c \). (b) Repeat for a 5.00-L flask. (c) Add 3.00 mol H2 to 10.00-L flask and recalculate.

Step 1: Define the Equilibrium Expression

The equilibrium expression for the reaction \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \) is given by:\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]Given \( K_c = 50.0 \) at \( 745\, \text{K} \).

Step 2: Set Initial Concentrations for 10.00-L Flask

For a 10.00-L flask, convert moles to concentrations:- \([\text{H}_2] = \frac{3.00 \, \text{mol}}{10.00 \, \text{L}} = 0.300 \, \text{M}\)- \([\text{I}_2] = \frac{1.00 \, \text{mol}}{10.00 \, \text{L}} = 0.100 \, \text{M}\)- \([\text{HI}] = 0 \, \text{M}\) initially.

Step 3: Apply Initial Changes and Define Equilibrium Concentrations

Assuming \( x \) is the change in concentration,\[ \text{At Equilibrium: } [\text{H}_2] = 0.300 - x, \quad [\text{I}_2] = 0.100 - x, \quad [\text{HI}] = 2x \]

Step 4: Substitute into Equilibrium Expression for 10.00-L Flask

Substitute equilibrium concentrations into the expression:\[ K_c = \frac{(2x)^2}{(0.300-x)(0.100-x)} = 50.0 \]Solve for \( x \).

Step 5: Solve for 'x'

Setting up the equation:\[ 4x^2 = 50.0((0.300-x)(0.100-x)) \]Simplify and solve this quadratic equation for \( x \).

Step 6: Calculate Number of Moles of HI at Equilibrium for 10.00-L Flask

Solve for \( x \) and substitute back:\[ [\text{HI}] = 2x \]Number of moles of HI in a 10.00-L flask is \( 10.00 \times 2x \).

Step 7: Repeat for 5.00-L Flask

Change initial concentrations for a 5.00-L flask:- \([\text{H}_2] = \frac{3.00 \, \text{mol}}{5.00 \, \text{L}} = 0.600 \, \text{M}\)- \([\text{I}_2] = \frac{1.00 \, \text{mol}}{5.00 \, \text{L}} = 0.200 \, \text{M}\)Follow Steps 3-6 similarly for solving \( x \) to find moles of HI.

Step 8: Calculate for Additional H2 in 10.00-L Flask

Add 3.00 mol more \( \text{H}_2 \) to the initial 3.00 mol, giving:- New \([\text{H}_2] = \frac{6.00 \, \text{mol}}{10.00 \, \text{L}} = 0.600 \, \text{M}\)Recalculate starting from Step 3 to find new equilibrium concentrations and final moles of HI.

Key Concepts

Equilibrium Constant

Chemical equilibrium is a dynamic state where the concentrations of reactants and products remain constant over time. The equilibrium constant, denoted as \(K_c\), quantifies this balance in a specific chemical reaction at a given temperature. For the reaction \(\mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{HI}(\mathrm{g})\), the equilibrium constant is given by the expression:

• \( K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \)

The value of \(K_c\) indicates the extent to which a reaction proceeds. A large \(K_c\) (much greater than 1) suggests that the equilibrium position favors the formation of products. Conversely, a very small \(K_c\) indicates the equilibrium lies towards the reactants. In this specific exercise, \(K_c\) is 50.0 at 745 K, illustrating that at equilibrium, the concentration of \(\text{HI}\) is significantly higher compared to \(\text{H}_2\) and \(\text{I}_2\). Understanding \(K_c\) helps predict how changes in concentration or pressure might shift the equilibrium.

Equilibrium Concentrations

Determining equilibrium concentrations involves understanding the initial concentrations and the changes that occur as the system reaches equilibrium. For the given reaction under certain conditions, we begin with known initial amounts of reactants and no products:

• \([\text{H}_2] = 0.300 \, \text{M}\)
• \([\text{I}_2] = 0.100 \, \text{M}\)
• \([\text{HI}] = 0 \, \text{M}\)

To solve for equilibrium concentrations, we use an ICE table (Initial, Change, Equilibrium) to track these amounts. The concentration change is modeled by an unknown \(x\), which denotes how much \(\text{H}_2\) and \(\text{I}_2\) are converted into \(\text{HI}\):

• \([\text{H}_2]_{eq} = 0.300 - x\)
• \([\text{I}_2]_{eq} = 0.100 - x\)
• \([\text{HI}]_{eq} = 2x\)

By substituting these equilibrium expressions into the \(K_c\) equation, we can solve for \(x\), thus determining the equilibrium concentrations of all species. It's crucial in chemistry to know how much product is formed and how much reactant remains.

Reaction Quotient

The reaction quotient \(Q\) provides a snapshot of the concentrations of the reactants and products at any point during a reaction, not just at equilibrium. It has the same mathematical form as the equilibrium constant \(K_c\):

• \( Q = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \)

When comparing \(Q\) to \(K_c\), we determine the direction in which the reaction must proceed to reach equilibrium:

• If \(Q < K_c\), the reaction will proceed in the forward direction (favoring product formation) until equilibrium is achieved.
• If \(Q > K_c\), the reaction will shift towards the reactants.
• If \(Q = K_c\), the reaction is already at equilibrium.

In our exercise, calculating \(Q\) using initial concentrations shows us if the system is at equilibrium or which way it needs to shift to reach it. Understanding \(Q\) helps predict how disturbances in concentration or pressure influence the equilibrium state.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that describes how a system at equilibrium will respond to disturbances. If an external change, like concentration alteration, pressure change, or temperature shift, is applied to a system at equilibrium, the system will adjust to counteract the change and restore a new equilibrium.When additional \(\text{H}_2\) was added to the system in one calculation exercise, it changed the initial concentration of \(\text{H}_2\). According to Le Chatelier's Principle:

• The system will adjust to minimize this change—by shifting the equilibrium to produce more \(\text{HI}\) from \(\text{H}_2\) and \(\text{I}_2\).
• This results in a new equilibrium state with different concentrations.

This principle is incredibly useful for predicting how changing conditions will affect the reaction, enabling chemists to manipulate the conditions to favor the formation of desired products in industrial processes.