Problem 8 - Chemistry and Chemical Reactivity

Question

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at a high temperature contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=3.77 \times 10^{-3} \mathrm{mol} / \mathrm{L},\left[\mathrm{O}_{2}\right]=\) \(4.30 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{SO}_{3}\right]=4.13 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) Calculate the equilibrium constant, \(K_{c}\), for the reaction. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) $$

Step-by-Step Solution

Answer

The equilibrium constant, \( K_c \), is approximately 279.

Step 1: Understand the Reaction Equation

The chemical reaction given is \( 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) \). The equilibrium constant expression, \( K_c \), for this reaction is \( K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]} \). We need to substitute the concentrations into this expression to find \( K_c \).

Step 2: Substitute the Concentrations

Insert the given concentration values into the \( K_c \) expression: \( [\mathrm{SO}_3] = 4.13 \times 10^{-3} \ \mathrm{mol/L} \), \( [\mathrm{SO}_2] = 3.77 \times 10^{-3} \ \mathrm{mol/L} \), and \( [\mathrm{O}_2] = 4.30 \times 10^{-3} \ \mathrm{mol/L} \).

Step 3: Calculate the Numerator

Calculate \([\mathrm{SO}_3]^2\): \( (4.13 \times 10^{-3})^2 = 1.70569 \times 10^{-5} \).

Step 4: Calculate the Denominator

Calculate \([\mathrm{SO}_2]^2 \times [\mathrm{O}_2]\):\( (3.77 \times 10^{-3})^2 \times (4.30 \times 10^{-3}) = 6.121029 \times 10^{-8} \).

Step 5: Compute the Equilibrium Constant

Substitute the values from Steps 3 and 4 into the \( K_c \) expression: \( K_c = \frac{1.70569 \times 10^{-5}}{6.121029 \times 10^{-8}} = 278.68 \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating state where a chemical reaction takes place in both directions at the same time. When a reaction reaches this state, the concentration of the reactants and products doesn't change anymore. It's like a balancing act between the forwards and backwards reaction.
When you look at a chemical equation like the one given in your exercise, you can understand the important role equilibrium plays. For the reaction \( 2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_3(\mathrm{g}) \), the system reaches a point where the formation of \( \mathrm{SO}_3 \) is equal to the decomposition back into \( \mathrm{SO}_2 \) and \( \mathrm{O}_2 \). This is what we call a dynamic balance, where reactants and products coexist in a fixed ratio.
It's important to note that equilibrium doesn't mean the chemical reaction has stopped. Instead, it's at a state of balance where reactants are converting to products and vice versa at equal rates.

Reaction Quotient

The reaction quotient, or \( Q \), is very much like a snapshot of a reaction at any given point. It gives us a sense of direction as to whether a reaction will reach equilibrium or not.
The mathematics behind the reaction quotient is similar to the equilibrium constant \( K_c \), calculated using the current concentrations of reactants and products. For the reaction \( 2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_3(\mathrm{g}) \), \( Q \) would be expressed as:
- \( Q = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2 [\mathrm{O}_2]} \).
By comparing \( Q \) to \( K_c \), we can determine the direction a reaction must proceed to reach equilibrium.

• If \( Q < K_c \), the reaction moves forward to form more products.
• If \( Q > K_c \), the reaction goes in reverse to create more reactants.
• If \( Q = K_c \), equilibrium has been achieved!

This comparison helps chemists predict and control processes in dynamic systems.

Concentration Calculations

Concentration calculations are crucial when you want to find the equilibrium constant, \( K_c \). Once you have the concentrations of all species involved, you can substitute these into the equilibrium expression.
For example, in the equation from your exercise: \( K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]} \), you have:

• \( [\mathrm{SO}_3] = 4.13 \times 10^{-3} \ \mathrm{mol/L} \)
• \( [\mathrm{SO}_2] = 3.77 \times 10^{-3} \ \mathrm{mol/L} \)
• \( [\mathrm{O}_2] = 4.30 \times 10^{-3} \ \mathrm{mol/L} \)

To find the equilibrium constant, calculate the squares and complements in the formula:

• Numerator: \( (4.13 \times 10^{-3})^2 \)
• Denominator: \( (3.77 \times 10^{-3})^2 \times (4.30 \times 10^{-3}) \)

By solving these, you get a clear view of how concentrations influence the equilibrium constant, enabling predictions about the reaction dynamics.

Dynamic Equilibrium

Dynamic equilibrium is a special stage of equilibrium in which reactions are continuously occurring, but without leading to a net change in the concentration of reactants and products. It's a condition of fascinating equilibrium in motion.
For the equation \( 2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_3(\mathrm{g}) \) being in dynamic equilibrium means that the rate of production of \( \mathrm{SO}_3 \) equals the rate at which \( \mathrm{SO}_3 \) breaks down back into \( \mathrm{SO}_2 \) and \( \mathrm{O}_2 \). It's like a dance where everyone is moving, but the positions ultimately remain unchanged.
Understanding this concept helps you appreciate why changing conditions, such as concentration, pressure, or temperature can shift the equilibrium. This is known as Le Chatelier's principle, a useful tool to predict how the system will adjust to re-establish equilibrium. By understanding dynamics, students can better grasp how real-world chemical processes operate and reach stability.