Question

An equilibrium mixture at a specific temperature is found to consist of \(1.2 \times 10^{-3} \mathrm{mol} / \mathrm{LHCl}\) , \(3.8 \times 10^{-4} \mathrm{mol} / \mathrm{L} \mathrm{O}_{2}, 5.8 \times 10^{-2} \mathrm{mol} / \mathrm{L} \mathrm{H}_{2} \mathrm{O},\) and \(5.8 \times 10^{-2} \mathrm{mol} / \mathrm{L} \mathrm{Cl}_{2}\) according to the following: \(\quad 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Cl}_{2}(g)\) Determine the value of the equilibrium constant for this system.

Short answer

The equilibrium constant, K_c, is 1.434 \times 10^{10}.

Step-by-step solution

Write down the equilibrium expression

For the reaction 4\text{HCl}(g) + \text{O}_2(g) \rightleftharpoons 2\text{H}_2\text{O}(g) + 2\text{Cl}_2(g),we need to write the equilibrium expression. The equilibrium constant expression, K_c, is given by:\[\text{K}_c = \frac{[\text{H}_2\text{O}]^2[\text{Cl}_2]^2}{[\text{HCl}]^4[\text{O}_2]}\]where the concentrations of \text{H}_2\text{O}, \text{Cl}_2, \text{HCl}, and \text{O}_2} are at equilibrium.

Substitute the equilibrium concentrations

Substitute the given concentrations into the equilibrium expression: \[\text{K}_c = \frac{(5.8 \times 10^{-2})^2 (5.8 \times 10^{-2})^2}{(1.2 \times 10^{-3})^4 (3.8 \times 10^{-4})}\]

Calculate the numerator

Calculate the product of the concentrations of the products: \[(5.8 \times 10^{-2})^2 = 3.364 \times 10^{-3}\] and since we are squaring it again for \text{Cl}_2, we get: \[(3.364 \times 10^{-3}) \times (3.364 \times 10^{-3}) = 1.131 \times 10^{-5}\]So, numerator is 1.131 \times 10^{-5}.

Calculate the denominator

Calculate the product of the concentrations of the reactants:\[(1.2 \times 10^{-3})^4 = 2.0736 \times 10^{-12}\] and \[3.8 \times 10^{-4} = 3.8 \times 10^{-4}\]Therefore, denominator is: \[(2.0736 \times 10^{-12}) \times (3.8 \times 10^{-4}) = 7.8808 \times 10^{-16}\]

Divide the numerator by the denominator

Now divide the numerator by the denominator to find \[K_c:\] \[K_c = \frac{1.131 \times 10^{-5}}{7.8808 \times 10^{-16}} = 1.434 \times 10^{10}\]

Key concepts

Chemical Equilibrium

In chemistry, reactions can reach a state where their forward and reverse reactions occur at the same rate. This state is known as chemical equilibrium. At this point, the concentrations of the reactants and products remain constant over time. However, it's important to note that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their ratios are stable. For example, in the reaction: 4HCl(g) + O₂(g) ⇌ 2H₂O(g) + 2Cl₂(g),we can analyze the equilibrium concentrations to understand the system's behavior more deeply.

Equilibrium Expression

The equilibrium expression is a mathematical representation that relates the concentrations of reactants and products at equilibrium to the equilibrium constant (K_c) for a given reaction. For the reaction 4HCl(g) + O₂(g) ⇌ 2H₂O(g) + 2Cl₂(g),the equilibrium constant expression is:\[K_c = \frac{[H_2O]^2[Cl_2]^2}{[HCl]^4[O_2]}\]Here, the square brackets represent the molar concentrations of each substance at equilibrium. The exponents correspond to the coefficients from the balanced chemical equation, ensuring a proportional relationship between reactants and products.

Reaction Quotient

The reaction quotient (Q) is an expression that uses the same formula as the equilibrium constant but with initial concentrations instead of equilibrium concentrations:\[Q = \frac{[H_2O]^2[Cl_2]^2}{[HCl]^4[O_2]}\]The main use of Q is to predict the direction in which a reaction will proceed to reach equilibrium. By comparing Q with K_c:

• If Q = K_c, the system is at equilibrium.
• If Q < K_c, the reaction will proceed forward to produce more products.
• If Q > K_c, the reaction will proceed in reverse to produce more reactants.

By knowing Q, we can predict how a reaction mixture will change over time.

Concentration Calculation

Calculating concentrations involves finding the amount of each reactant and product present at equilibrium. Given the reaction: 4HCl(g) + O₂(g) ⇌ 2H₂O(g) + 2Cl₂(g),with equilibrium concentrations of 1.2 × 10⁻³ mol/L HCl, 3.8 × 10⁻⁴ mol/L O₂, 5.8 × 10⁻² mol/L H₂O, and 5.8 × 10⁻² mol/L Cl₂,we substitute these values into the equilibrium expression to solve for K_c. The steps are:
1. Substitute the equilibrium concentrations:\(K_c = \frac{(5.8 \times 10^{-2})^2 (5.8 \times 10^{-2})^2}{(1.2 \times 10^{-3})^4 (3.8 \times 10^{-4})}\)2. Calculate the numerator and denominator separately:\((5.8 \times 10^{-2})^4 = 1.131 \times 10^{-5}\)and\((1.2 \times 10^{-3})^4 \times 3.8 \times 10^{-4} = 7.8808 \times 10^{-16}\)3. Finally, divide the numerator by the denominator to find \(K_c = 1.434 \times 10^{10}\).

Holt Modern Chemistry

The textbook 'Holt Modern Chemistry' is a comprehensive resource for understanding various chemistry concepts, including chemical equilibrium. It provides detailed explanations, example problems, and practice exercises that help students grasp difficult topics. When studying equilibrium constant calculations, Holt Modern Chemistry will often take you through the process step-by-step. This includes defining important terms, explaining the logic behind the equilibrium expression, and guiding you through sample calculations, just like we've done above.
Using the textbook as a support tool, students can deepen their understanding of subjects like equilibrium constant calculations and confidently approach similar problems in their coursework.