Problem 35 - Chemistry The Central Science In Si Units

Question

A mixture of 0.140 mol of \(\mathrm{NO}, 0.060 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and 0.260 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.0-L vessel at \(330 \mathrm{~K}\). Assume that the following equilibrium is established: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At equilibrium \(\left[\mathrm{H}_{2}\right]=0.010 \mathrm{M} .(\mathbf{a})\) Calculate the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) (b) Calculate \(K_{c}\).

Step-by-Step Solution

Answer

Equilibrium concentrations: [NO] = 0.050 M, [N2] = 0.010 M, [H2O] = 0.150 M. The equilibrium constant \(K_c\) is 9000.

Step 1: Calculate Initial Concentrations

First, calculate the initial concentrations of NO, H2, and H2O by using the formula:\[ [A] = \frac{{\text{{moles of A}}}}{{\text{{volume in L}}}} \]- For NO: \([\text{NO}]_0 = \frac{0.140 \text{ mol}}{2.0 \text{ L}} = 0.070 \text{ M}\)- For H2: \([\text{H}_2]_0 = \frac{0.060 \text{ mol}}{2.0 \text{ L}} = 0.030 \text{ M}\)- For H2O: \([\text{H}_2O]_0 = \frac{0.260 \text{ mol}}{2.0 \text{ L}} = 0.130 \text{ M}\)

Step 2: Define Changes and Set Up ICE Table

Using the initial concentrations, set up an ICE (Initial, Change, Equilibrium) table to find the changes in concentrations. Define the change in concentration of \([\text{H}_2]\) as \(-x\). Since \([\text{H}_2]\) at equilibrium is given as 0.010 M:\[\Delta [\text{H}_2] = 0.030 - 0.010 = -x = -0.020 \text{ M}\]This results in:- Change for \([\text{NO}]\): \(-x = -0.020 \text{ M}\)- Change for \([\text{N}_2]\): \(+\frac{x}{2} = +0.010 \text{ M}\)- Change for \([\text{H}_2O]\): \(+x = +0.020 \text{ M}\)

Step 3: Calculate Equilibrium Concentrations

Use the changes to calculate equilibrium concentrations:- For NO: \([\text{NO}]_e = [\text{NO}]_0 - x = 0.070 - 0.020 = 0.050 \text{ M}\)- For N2: \([\text{N}_2]_e = 0 + 0.010 = 0.010 \text{ M}\)- For H2O: \([\text{H}_2O]_e = [\text{H}_2O]_0 + x = 0.130 + 0.020 = 0.150 \text{ M}\)

Step 4: Calculate Equilibrium Constant \(K_c\)

The equilibrium expression for the reaction is given by:\[ K_c = \frac{[\text{N}_2][\text{H}_2O]^2}{[\text{NO}]^2[\text{H}_2]^2} \]Substitute in the equilibrium concentrations:\[K_c = \frac{(0.010)(0.150)^2}{(0.050)^2(0.010)^2} \]Calculate to find:\[K_c = \frac{0.00225}{0.00000025} = 9000\]

Key Concepts

ICE Table

The ICE table is a practical tool used to help visualize and organize information for chemical equilibrium reactions. It stands for Initial, Change, and Equilibrium, which are the key phases to consider when solving these types of problems. The table is structured by listing the reactants and products of the chemical reaction and plotting their concentrations over these three phases.

Initially, you populate the table with the initial concentrations of each substance involved in the reaction. In our example, using a volume of 2.0 L, we calculated the initial concentration for \(\text{NO}\) as 0.070 M, for \(\text{H}_2\) as 0.030 M, and for \(\text{H}_2\text{O}\) as 0.130 M.

• Initial: The first row in an ICE table contains initial concentrations before any reaction has occured.
• Change: This row indicates the change in concentration for each substance as the system transitions to equilibrium.
• Equilibrium: This row holds the concentrations of substances once equilibrium is reached.

The 'Change' portion uses variable expressions (such as \(-x\) or \(+x\)) to dictate how each concentration is expected to change as the reaction moves to equilibrium based on stoichiometry. For instance, if the change in \(\text{H}_2\) is \(-0.020 \, \text{M}\), this means it decreases by this amount to form the products. Once changes are known, you calculate the equilibrium concentrations by combining initial concentrations with these changes.

Equilibrium Constant (Kc)

The Equilibrium Constant, denoted as \(K_c\), is a numerical value that represents the ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients from the balanced chemical equation, at equilibrium. This constant provides insight into the extent of a reaction; a large \(K_c\) means the reaction heavily favors products, while a small \(K_c\) indicates it favors reactants.

For the reaction \(2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{N}_2(g) + 2 \text{H}_2\text{O}(g)\), the equilibrium expression for \(K_c\) is structured like this:
\[ K_c = \frac{[\text{N}_2][\text{H}_2\text{O}]^2}{[\text{NO}]^2[\text{H}_2]^2} \]
The concentration values are those we calculated at equilibrium. Substituting known values yields:
\[ K_c = \frac{(0.010 \text{ M})(0.150 \text{ M})^2}{(0.050 \text{ M})^2(0.010 \text{ M})^2} \]
This calculation returns \(K_c = 9000\), which reflects that the reaction significantly shifts towards forming products. Understanding \(K_c\) helps predict how changes in conditions (like concentration, pressure, or temperature) might affect the system.

Initial Concentrations Calculation

Calculating initial concentrations is a necessary step in chemistry, especially within equilibrium problems. Initial concentrations provide the starting point for how much of each reactant and product is present before the reaction reaches equilibrium.

To find these values, use the formula:
\[ [A] = \frac{\text{moles of A}}{\text{volume in L}} \]
In our example, the initial concentrations are calculated for each substance as follows:

• \(\text{NO}\): \([\text{NO}]_0 = \frac{0.140 \, \text{mol}}{2.0 \, \text{L}} = 0.070 \, \text{M}\)
• \(\text{H}_2\): \([\text{H}_2]_0 = \frac{0.060 \, \text{mol}}{2.0 \, \text{L}} = 0.030 \, \text{M}\)
• \(\text{H}_2\text{O}\): \([\text{H}_2\text{O}]_0 = \frac{0.260 \, \text{mol}}{2.0 \, \text{L}} = 0.130 \, \text{M}\)

Understanding and obtaining the correct initial concentrations set the correct path for the subsequent steps of analyzing the equilibrium system. They help in determining the extent to which changes will occur as equilibrium is approached, ultimately impacting the resulting equilibrium concentrations.