Question

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 3.00-L container at \(395^{\circ} \mathrm{C},\) the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) . \quad\) If \(\quad K_{c}=0.802\) what are the concentrations of each substance in the equilibrium mixture?

Short answer

At equilibrium, the concentrations of the species are: \([\mathrm{CO}_{2}] = 0.277\,\mathrm{M}\), \([\mathrm{H}_{2}] = 0.277\,\mathrm{M}\), \([\mathrm{CO}] = 0.223\,\mathrm{M}\), and \([\mathrm{H}_{2}\mathrm{O}] = 0.223\,\mathrm{M}\).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation is given as: \(\mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(g)\)

Set up an ICE table

An ICE (Initial, Change, Equilibrium) table helps track the changes in species concentrations throughout the reaction process. Initial concentrations: \(\frac{1.50\,\mathrm{mol}}{3.00\,\mathrm{L}} = 0.500\,\mathrm{M}\) for \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\) Initial concentrations of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\mathrm{O}\) are 0 M since they are not given in the problem statement. ``` CO2 H2 CO H2O Initial 0.500 0.500 0 0 Change -x -x +x +x Final 0.500-x 0.500-x x x ```

Write the equilibrium constant expression

\(K_c = \frac{[\mathrm{CO}] [\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CO}_{2}] [\mathrm{H}_{2}]}\)

Substitute ICE table values into the equilibrium expression

We substitute the equilibrium concentrations from the ICE table into the \(K_c\) expression: \(0.802 = \frac{x^2}{(0.500-x)^2}\)

Solve for x

To solve for x, we first notice that the denominator and numerator have the same exponent. Thus, we can simplify the equation: \(0.802 = \frac{x}{0.500-x}\) Now, multiply both sides by \((0.500-x)\): \(0.802(0.500-x) = x\) Distribute the 0.802 on the left-hand side: \(0.401 - 0.802x = x\) Combine the x terms and solve for x: \(x = \frac{0.401}{1.802} \approx 0.223\,\mathrm{M}\)

Calculate the equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations of all species: \([\mathrm{CO}]_{eq} = [\mathrm{H}_{2}\mathrm{O}]_{eq} = x = 0.223\,\mathrm{M}\) \([\mathrm{CO}_{2}]_{eq} = [\mathrm{H}_{2}]_{eq} = 0.500 - x = 0.500 - 0.223 = 0.277\,\mathrm{M}\) So, at equilibrium, the concentrations of the species are: \([\mathrm{CO}_{2}] = 0.277\,\mathrm{M}\) \([\mathrm{H}_{2}] = 0.277\,\mathrm{M}\) \([\mathrm{CO}] = 0.223\,\mathrm{M}\) \([\mathrm{H}_{2}{\mathrm{O}}] = 0.223\,\mathrm{M}\)

Key concepts

ICE table

An ICE table is a valuable tool for understanding chemical reactions and their progress towards equilibrium. The term "ICE" stands for Initial, Change, and Equilibrium, reflecting the different stages of a reaction.
It's used to track the molar concentrations of reactants and products as the reaction progresses. Initially, we list what we know:

• **Initial** concentrations: We begin by calculating the initial molarity of each reactant, using the formula: Molarity = moles/volume (in liters). For example, in our exercise, both the CO2 and H2 start with a concentration of 0.500 M.
• **Change** in concentration: As the reaction proceeds, reactants are consumed, and products are formed. We represent these changes using a variable, often denoted as `x`. As CO2 and H2 react to form CO and H2O, they decrease by the amount `x`, while the products increase by `x`.
• **Equilibrium** concentration: Finally, we express the equilibrium concentrations in terms of `x`. These are used in further calculations to find how much of each substance is present when the reaction reaches equilibrium.

This structured approach simplifies the often complex changes that occur in a reaction, making it easier to solve equilibrium problems.

equilibrium constant (Kc)

The equilibrium constant, or \(K_c\), is a key concept in chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation, at equilibrium.
In essence, \(K_c\) tells us how far a reaction proceeds before reaching equilibrium. If \(K_c > 1\), the products are favored at equilibrium, meaning more of the product is present than the reactant. Conversely, if \(K_c < 1\), the reactants are favored.
In mathematical terms, for a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression is:

• \(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

Here, the brackets denote molar concentrations.
In the exercise, the equilibrium constant \(K_c = 0.802\) indicates that, at equilibrium, there is a moderate amount of reaction towards products, but the reaction does not heavily favor them. This value is critical for determining concentrations of substances at equilibrium using the ICE table values.

reaction stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It provides a blueprint of how much reactant is needed to form a particular amount of product.
In the exercise's reaction:\[\mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(g)\]each molecule of CO2 reacts with one molecule of H2 to produce one molecule of CO and one of H2O.
Stoichiometry is essential when filling in an ICE table:

• We assign changes in concentration based on these stoichiometric coefficients. For example, for each mole of CO2 consumed, an equal amount of CO is produced.
• This one-to-one relationship simplifies calculations, but in reactions with different coefficients, it’s crucial to correctly scale changes according to these ratios.

Reaction stoichiometry not only ensures that your ICE table is set up accurately, but it also provides the foundation for calculating equilibrium concentrations correctly. Understanding these stoichiometric relationships is key to mastering chemical equilibrium problems.