Question

\(1.00 \mathrm{g}\) each of \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{H}_{2}\) are sealed in a \(1.41 \mathrm{L}\) vessel and brought to equilibrium at 600 K. How many grams of \(\mathrm{CO}_{2}\) will be present in the equilibrium mixture? $$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=23.2$$

Short answer

The mass of \(\mathrm{CO}_2\) in the equilibrium mixture cannot be determined without further calculations and it will depend on the computed positive value of x in step 4.

Step-by-step solution

Calculating Initial Moles

Calculate the initial moles of the reactants and products. Use the molar mass of the substances to convert the grams to moles. For \(\mathrm{CO}\): \(\frac{1.00 \, \mathrm{g}}{28.01 \, \mathrm{g/mol}} \approx 0.036 \, \mathrm{mol}\)For \(\mathrm{H_2O}\): \(\frac{1.00 \, \mathrm{g}}{18.02 \, \mathrm{g/mol}} \approx 0.055 \, \mathrm{mol}\)For \(\mathrm{H_2}\) and \(\mathrm{CO}_2\): Initially, these are 0 as the reaction hasn't started yet.

Write the ICE Table

Set up an ICE (Initial, Change, Equilibrium) table to determine the changes in the moles at equilibrium. The reaction is \(\mathrm{CO}\) + \(\mathrm{H_2O}\) ⇌ \(\mathrm{CO}_2\) + \(\mathrm{H_2}\). The change in moles during the reaction is represented as x. At equilibrium, there is -x change for reactants and +x change for the products.

Use the Equilibrium Constant

Use the equilibrium constant expression to write the equation. \(K_c\) equals the ratio of the product concentrations to the reactant concentrations, each raised to the power equal to their coefficients in the balanced equation. Given \(K_c = 23.2\), we can write: \[23.2 = \frac{[(0+x)(0+x)]}{[(0.036-x)(0.055-x)]}\]

Solve for x

This is a quadratic equation, rewrite it as one and solve for x. You will get two values. Choose the value that makes physical sense (moles cannot be negative).

Find the mass of \(\mathrm{CO}_2\)

Substitute x into the equilibrium moles of \(\mathrm{CO}_2\) to get the moles of \(\mathrm{CO}_2\) at equilibrium. Convert this amount into grams by using the molar mass of \(\mathrm{CO}_2\) (about 44.01 g/mol). The result will give you the mass of \(\mathrm{CO}_2\) at equilibrium.

Key concepts

ICE Table

The ICE Table is an essential tool in solving equilibrium problems in chemistry. ICE stands for Initial, Change, and Equilibrium. It helps you visualize the transition of reactants to products over time.
In the ICE Table, the first row lists the initial concentrations or moles of reactants and products before the reaction starts. The second row represents the change in concentration or moles as the reaction proceeds to equilibrium; usually denoted with the variable 'x' to signify the shift. Finally, the third row shows the equilibrium concentrations or moles once the system is at equilibrium.
The ICE Table simplifies the tracking of changes in amounts of each substance, making it easier to set up the expressions required for further calculations like the Equilibrium Constant. Understanding this table is key to predicting the behavior of reactants and products in a chemical equilibrium system.

Equilibrium Constant

The Equilibrium Constant, designated as \(K_c\), quantifies the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. This ratio is raised to the power of their stoichiometric coefficients.
For the reaction \(\text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)}\) the equilibrium constant \(K_c\) is given by:\[K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}\]
In this problem, \(K_c = 23.2\), so the products are favored in the equilibrium mixture. Knowing \(K_c\) allows you to set up an equation using the ICE Table values to solve for x, the change in molar quantities, and predict concentrations at equilibrium.

Quadratic Equation

A Quadratic Equation in an equilibrium calculation arises from the expression \(K_c\). Let's illustrate using the equilibrium constant from above:
Incorporating changes using \(x\) from the ICE Table:\[23.2 = \frac{[(0 + x)(0 + x)]}{[(0.036 - x)(0.055 - x)]}\]
This equation simplifies to a quadratic form: \(ax^2 + bx + c = 0\). Solving this involves techniques like factoring, completing the square, or the quadratic formula. The quadratic equation often yields two solutions, but only the one yielding non-negative concentrations makes physical sense.
Solving quadratic equations in chemical equilibrium can seem challenging, but once you set it up using appropriate changes from the ICE Table, it becomes manageable.

Molar Mass Calculation

Calculating molar mass is critical to convert between grams and moles, a necessity in equilibrium problems. Molar mass is the mass of one mole of a substance and is expressed in \(g/mol\). For example, the molar mass of \(\text{CO}\) is about 28.01 \(g/mol\), \(\text{H}_2\text{O}\) is about 18.02 \(g/mol\), and \(\text{CO}_2\) is approximately 44.01 \(g/mol\).
To calculate the initial moles of a substance, such as \(\text{CO}\), use:\[moles = \frac{\text{mass}}{\text{molar mass}}\] For \(1.00 \text{g}\) of \(\text{CO}\), this calculation yields around \(0.036\text{ moles}\).
Converting moles back to grams at the end helps determine how much of a product, like \(\text{CO}_2\), is present at equilibrium.