Question

To study the following reaction at \(20^{\circ} \mathrm{C}\), \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HNO}_{2}(g)\) a mixture of \(\mathrm{NO}(g), \mathrm{NO}_{2}(g)\), and \(\mathrm{H}_{2} \mathrm{O}(g)\) was prepared in a \(10.0 \mathrm{~L}\) glass bulb. For \(\mathrm{NO}, \mathrm{NO}_{2},\) and \(\mathrm{HNO}_{2},\) the initial concentrations were as follows: \([\mathrm{NO}]=\left[\mathrm{NO}_{2}\right]=\) \(2.59 \times 10^{-3} M\) and \(\left[\mathrm{HNO}_{2}\right]=0 M .\) The initial partial pressure of \(\mathrm{H}_{2} \mathrm{O}(g)\) was 17.5 torr. When equilibrium was reached, the \(\mathrm{HNO}_{2}\) concentration was \(4.0 \times 10^{-4} \mathrm{M}\) Calculate the equilibrium constant, \(K_{\mathrm{c}},\) for this reaction.

Short answer

Based on the initial and equilibrium concentrations, use stoichiometry to find the changes in concentration, set up the ICE table, calculate the equilibrium concentrations for NO, NO2 and H2O, then substitute into the equilibrium expression to solve for \(K_{\mathrm{c}}\).

Step-by-step solution

Write the Equilibrium Expression

For the given reaction \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HNO}_{2}(g)\), write the equilibrium expression for \(K_{\mathrm{c}}\) which is the ratio of the product's concentrations to the reactants' concentrations, each raised to the power of its coefficient in the balanced equation. \[ K_{\mathrm{c}} = \frac{[\mathrm{HNO}_{2}]^2}{[\mathrm{NO}][\mathrm{NO}_{2}][\mathrm{H}_{2} \mathrm{O}]} \]

Calculate The Change in Concentrations

From the initial concentrations and the equilibrium concentration of \(\mathrm{HNO}_{2}\), determine the changes in concentrations of the reactants and the product. Since 2 moles of \(\mathrm{HNO}_{2}\) are produced, the change for \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) must be half that amount because they are in a 1:1 ratio with the product (but subtracted because they get consumed).

Set Up the ICE Table

Using an ICE (Initial, Change, Equilibrium) table, we set up the initial concentrations, changes, and the equilibrium concentrations for all species in the reaction. For water, the pressure must be converted into concentration using the ideal gas equation \(PV=nRT\), approximating that 1 mole of gas occupies \(22.4L\) at room temperature and pressure.

Calculate the Equilibrium Concentrations of Reactants

Given the stoichiometry of the reaction, the initial concentrations of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\), and the equilibrium concentration of \(\mathrm{HNO}_{2}\), we can calculate the equilibrium concentrations of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). \( [\mathrm{NO}]_{eq} = [\mathrm{NO}_{2}]_{eq} = 2.59 \times 10^{-3} M - 2 \times (4.0 \times 10^{-4} M) \)

Calculate the Equilibrium Concentration of H2O

The partial pressure of \(\mathrm{H}_{2} \mathrm{O}(g)\) in torr can be converted to atm by dividing by 760. Then, using the ideal gas law \(PV = nRT\), we can find the concentration in moles per liter (M).

Calculate the Equilibrium Constant, Kc

Substitute the equilibrium concentrations into the equilibrium expression (from Step 1) and solve for \(K_{\mathrm{c}}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium represents a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the concentrations of the reactants and products remain constant over time, though they are not necessarily equal.

Equilibrium can only happen in a closed system and is dynamic, meaning the reactions are still occurring, but there's no net change in concentration. It's important to understand that reaching equilibrium doesn't mean the reactants and products are in a 1:1 ratio, but rather that they have stabilized at particular concentrations that depend on the reaction conditions and the equilibrium constant, denoted as \( K \).

In the given exercise, the equilibrium constant \( K_{\text{c}} \) needs to be calculated for the reaction involving nitrogen monoxide, nitrogen dioxide, and water vapor producing nitrous acid. At equilibrium, although the concentrations do not change, understanding the equilibrium expression and the constants involved allows chemists to predict how a system will react to changes in conditions, following Le Chatelier's principle.

ICE Table

The ICE table is a systematic approach to handle the quantitative aspects of chemical equilibrium problems. ICE stands for Initial, Change, and Equilibrium, which represent the stages of the reaction.

To use the ICE table effectively, one should start by writing the balanced chemical equation and then under it, list the initial molar concentrations of the reactants and products. 'Change' represents the changes in concentrations as the system moves toward equilibrium. This is often unknown and can be expressed in terms of 'x', based on the stoichiometry of the reaction. Finally, 'Equilibrium' involves writing expressions for the expected concentrations of reactants and products at equilibrium, incorporating the changes from the initial state.

In practice, once you determine the change in concentration for one substance at equilibrium (often found experimentally or given in the problem), you can use the stoichiometry of the reaction to find the changes for the other substances. Then, these changes are applied to the initial concentrations to find the equilibrium concentrations for all substances involved.

Reaction Stoichiometry

Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It tells us the ratios in which chemicals react with one another and the ratios in which they produce products. Stoichiometry is critical when setting up equilibrium problems because these ratios determine the 'Change' row in an ICE table.

For the reaction \( \text{NO}(g) + \text{NO}_2(g) + \text{H}_2 \text{O}(g) \rightleftharpoons 2 \text{HNO}_2(g) \), the stoichiometry highlights that one mole of \( \text{NO} \) reacts with one mole of \( \text{NO}_2 \) and one mole of \( \text{H}_2 \text{O} \) to produce two moles of \( \text{HNO}_2 \). This means that for every one mole of \( \text{HNO}_2 \) formed, half a mole of both \( \text{NO} \) and \( \text{NO}_2 \) are consumed.

When we solve equilibrium problems, using stoichiometry allows us to express the concentration changes of all reactants and products in terms of a single variable, significantly simplifying the process of finding equilibrium concentrations for each species.