Problem 8 - Chemistry and Chemical Reactivity

Question

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at \(1000 \mathrm{K}\) contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]\) $$ =3.77 \times 10^{-3} \mathrm{mol} / \mathrm{L},\left[\mathrm{O}_{2}\right]=4.30 \times 10^{-3} \mathrm{mol} / \mathrm{L}, \text { and } $$ \(\left[\mathrm{SO}_{3}\right]=4.13 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Calculate the equilibrium constant, \(K\), 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, \( K \), is 278.

Step 1: Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the given reaction is determined by the formula \( K = \frac{\left[ \mathrm{SO}_3 \right]^2}{\left[ \mathrm{SO}_2 \right]^2 \times \left[ \mathrm{O}_2 \right]} \). This is derived from the balanced chemical reaction \( 2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g}) \), where the equilibrium concentrations are raised to the power of their coefficients in the balanced equation.

Step 2: Substitute Concentrations into the Expression

Substitute the given concentrations into the equilibrium expression: \[ K = \frac{\left(4.13 \times 10^{-3}\right)^2}{\left(3.77 \times 10^{-3}\right)^2 \times 4.30 \times 10^{-3}} \].

Step 3: Calculate the Numerator and the Denominator

Calculate the numerator: \( \left(4.13 \times 10^{-3}\right)^2 = 1.706 \times 10^{-5} \). Next, calculate the denominator: \( \left(3.77 \times 10^{-3}\right)^2 \times 4.30 \times 10^{-3} = 6.128 \times 10^{-8} \).

Step 4: Calculate the Equilibrium Constant

Divide the numerator by the denominator to find \( K \): \[ K = \frac{1.706 \times 10^{-5}}{6.128 \times 10^{-8}} = 278 \].

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating state in any reversible chemical reaction. In this state, the rate at which the reactants convert into products is exactly equal to the rate at which the products revert back into reactants. Thus, the concentrations of all components remain constant over time.
In the case of the reaction involving sulfur dioxide (\(\mathrm{SO}_2\)), oxygen (\(\mathrm{O}_2\)), and sulfur trioxide (\(\mathrm{SO}_3\)), a chemical equilibrium is reached when these gases achieve concentrations that do not change with time, demonstrating a balance of forward and reverse reactions.
The equilibrium constant, \(K\), provides a quantitative measure of the position of equilibrium. For the reaction \(2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g})\), the expression for \(K\) is: \[ K = \frac{\left[ \mathrm{SO}_3 \right]^2}{\left[ \mathrm{SO}_2 \right]^2 \times \left[ \mathrm{O}_2 \right]} \]
By substituting the equilibrium concentrations into the expression, we can determine \(K\) and understand the extent to which the reaction favors either reactants or products at the given temperature of 1000 K.

Reaction Kinetics

Reaction kinetics is the study of the speeds of chemical processes and the factors that affect these speeds. Understanding kinetics is vital for predicting how quickly a reaction reaches equilibrium.
The reaction of \(\mathrm{SO}_2\), \(\mathrm{O}_2\), and \(\mathrm{SO}_3\) displays its kinetics by the balanced chemical equation: \(2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g})\). Kinetics explores how different factors such as temperature, pressure, and concentration influence the reaction speed.
At 1000 K, the high temperature most likely accelerates the forward reaction, increasing the rate at which \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) form \(\mathrm{SO}_3\). However, once equilibrium is achieved, the rates of both forward and backward reactions stabilize.
Observing the concentration changes over time can give insights into the mechanisms and rate-determining steps that govern the reaction pathway, providing a link to how concentrations stabilize at equilibrium.

Concentration Calculations

Calculating concentrations at equilibrium provides insights into the dynamics of chemical reactions. To perform these calculations, we use the equilibrium constant formula, often involving the stoichiometry of the balanced chemical equation.
For the given reaction, \(K\) is expressed as \[ K = \frac{\left[ \mathrm{SO}_3 \right]^2}{\left[ \mathrm{SO}_2 \right]^2 \times \left[ \mathrm{O}_2 \right]} \] based on the stoichiometry from the balanced equation.
Given the concentrations \([\mathrm{SO}_2] = 3.77 \times 10^{-3} \mathrm{mol}/\mathrm{L}\), \([\mathrm{O}_2] = 4.30 \times 10^{-3} \mathrm{mol}/\mathrm{L}\), and \([\mathrm{SO}_3] = 4.13 \times 10^{-3} \mathrm{mol}/\mathrm{L}\), we substitute these values into the formula:
\[ K = \frac{(4.13 \times 10^{-3})^2}{(3.77 \times 10^{-3})^2 \times (4.30 \times 10^{-3})} \]
This calculation involves: - Calculating the numerator: \((4.13 \times 10^{-3})^2\) - Calculating the denominator: \((3.77 \times 10^{-3})^2 \times 4.30 \times 10^{-3}\)
These concentration calculations are paramount in determining \(K\), illustrating how equilibrium shifts in response to concentration changes, hence offering a quantitative comparison of reaction tendencies.