Problem 30 - Chemistry: An Atoms-Focused Approach

Question

A 100 mL reaction vessel initially contains \(2.60 \times 10^{-2} \mathrm{mol} \mathrm{NO}\) and \(1.30 \times 10^{-2} \mathrm{mol} \mathrm{H}_{2}, \mathrm{At}\) cquilibrium, the concentration of NO in the vesscl is \(0.161 M .\) What is the value of \(K_{\mathrm{c}}\) for the following reaction? $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$

Step-by-Step Solution

Answer

In this problem, we are given the initial amounts of NO and H2, and the equilibrium concentration of NO. We are asked to find the equilibrium constant, Kc, for the given reaction. Following the given steps, we find the initial concentrations of all species, determine the changes in concentrations at equilibrium, calculate the equilibrium concentrations of all species, and finally, calculate Kc. The equilibrium constant for this reaction is found to be approximately 9.76.

Step 1: Calculate the initial concentrations of all species

Given that the 100 mL reaction vessel initially contains 2.60 × 10-2 mol NO and 1.30 x 10-2 mol H2, we can calculate their initial concentrations by dividing the initial moles by the volume of the reaction vessel (in liters). Since there is no mention of N2 and H2O at the beginning, we will assume their initial concentrations to be 0. Initial concentration of NO = \((2.6\times10^{-2}\,\text{mol})\div(0.1\,\text{L})=0.26\,\text{M}\) Initial concentration of H2 = \((1.3\times10^{-2}\,\text{mol})\div(0.1\,\text{L})=0.13\,\text{M}\) Initial concentration of N2 = 0 M Initial concentration of H2O = 0 M Note that the initial concentration of NO and H2 are equal since their stoichiometric coefficients are also equal.

Step 2: Determine the change in concentrations at equilibrium

We know the equilibrium concentration of NO is 0.161 M. Since NO is a reactant, its concentration decreases at equilibrium. The change in NO concentration can be represented as: \(\text{Final(NO)} - \text{Initial(NO)}= -0.1x\) We can plug in the values to calculate the value of x: \(0.161\,\text{M} - 0.26\,\text{M}=-0.1x\) \(x=0.99\) Now, we can calculate the change in concentrations for all other species using stoichiometry. For every 2 moles of NO consumed, 2 moles of H2 are also consumed, 2 moles of H2O are produced, and 1 mole of N2 is produced. Change in H2 concentration = \(-0.1x = -0.099\,\text{M}\) Change in H2O concentration = \(0.1x=0.099\,\text{M}\) Change in N2 concentration = \(0.05x=0.0495\,\text{M}\)

Step 3: Calculate the equilibrium concentrations of all species

Now, we will use the initial concentrations and the change in concentrations to find the equilibrium concentrations of all species: Equilibrium concentration of NO = Initial(NO) + Change in NO = \(0.26\,\text{M} - 0.099\,\text{M}=0.161\,\text{M}\) Equilibrium concentration of H2 = Initial(H2) + Change in H2 = \(0.13\,\text{M} - 0.099\,\text{M}=0.031\,\text{M}\) Equilibrium concentration of H2O = Initial(H2O) + Change in H2O = \(0\,\text{M} + 0.099\,\text{M}=0.099\,\text{M}\) Equilibrium concentration of N2 = Initial(N2) + Change in N2 = \(0\,\text{M} + 0.0495\,\text{M}=0.0495\,\text{M}\)

Step 4: Calculate the value of Kc

We will use the equilibrium concentrations of all species in the reaction to calculate the value of Kc: $$ K_c=\frac{[\mathrm{H}_{2}\mathrm{O}]^2[\mathrm{N}_{2}]}{[\mathrm{H}_{2}]^2[\mathrm{NO}]^2} $$ Substituting the equilibrium concentrations: $$ K_c=\frac{(0.099)^2 (0.0495)}{(0.031)^2 (0.161)^2} $$ Calculate the value of Kc: $$ K_c\approx 9.76 $$ The value of Kc for the given reaction is approximately 9.76.

Key Concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant, represented as \(K_c\), conveys the ratio of product concentrations to reactant concentrations at equilibrium. This constant is crucial because it provides insight into the balance between reactants and products in a reversible reaction. For the reaction \(2\text{H}_2(g) + 2\text{NO}(g) \rightleftharpoons 2\text{H}_2\text{O}(g) + \text{N}_2(g)\), the expression for \(K_c\) is derived from the balanced chemical equation. The formula becomes:

• \(K_c = \frac{[\text{H}_2\text{O}]^2[\text{N}_2]}{[\text{H}_2]^2[\text{NO}]^2}\)

Here, each concentration is raised to the power corresponding to its stoichiometric coefficient in the balanced equation. Calculating \(K_c\) involves substituting the equilibrium concentrations into this equation. It tells us whether products or reactants are favored at equilibrium.

Reaction Stoichiometry

Stoichiometry in chemical equilibrium reactions helps us understand how the quantities of reactants and products relate to each other. In our reaction, the stoichiometric coefficients are

• 2 for \(\text{H}_2\)
• 2 for \(\text{NO}\)
• 2 for \(\text{H}_2\text{O}\)
• 1 for \(\text{N}_2\)

These coefficients guide converting moles and concentrations between different chemical species. For every 2 moles of \(\text{NO}\) that react, 2 moles of \(\text{H}_2\) also react, and as products, 2 moles of \(\text{H}_2\text{O}\) and 1 mole of \(\text{N}_2\) are formed. This relationship is essential in determining how much each species will increase or decrease as equilibrium is reached.

Initial Concentrations

Initial concentrations are foundational when determining how equilibrium shifts. At the outset of our example reaction, the given amounts were

• \(2.60 \times 10^{-2}\,\text{mol}\) of \(\text{NO}\)
• \(1.30 \times 10^{-2}\,\text{mol}\) of \(\text{H}_2\)

In the 100 mL vessel, this translates to initial concentrations of

• \(0.26\,\text{M}\) for \(\text{NO}\)
• \(0.13\,\text{M}\) for \(\text{H}_2\)

There were no initial \(\text{N}_2\) or \(\text{H}_2\text{O}\); thus their concentrations started at \(0\,\text{M}\). Correctly establishing these initial concentrations allows us to predict and calculate the changes that occur as the reaction moves to equilibrium.

Equilibrium Concentrations

Equilibrium concentrations tell us how much of each substance is present when the reaction reaches equilibrium. Using both initial concentrations and stoichiometry, we determine these values. In our example:

• The equilibrium concentration of \(\text{NO}\) is found to be \(0.161\,\text{M}\).
• For \(\text{H}_2\), the equilibrium concentration is calculated as \(0.031\,\text{M}\).
• The equilibrium concentration of \(\text{H}_2\text{O}\) is \(0.099\,\text{M}\).
• For \(\text{N}_2\), it's \(0.0495\,\text{M}\).

These values result from the reaction adjusting to equilibrium, where concentrations stop changing. By using the initial amounts and the changes determined through stoichiometry, we can reliably calculate these equilibrium concentrations, which are essential for determining \(K_c\).