Question

A mixture of \(0.10 \mathrm{~mol}\) of \(\mathrm{NO}, 0.050 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(1.0\) -L vessel at \(300 \mathrm{~K}\). 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 \([\mathrm{NO}]=0.062 \mathrm{M} .\) (a) Calculatethe equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\). (b) Calculate \(K_{\mathrm{c}}\)

Short answer

At equilibrium, the concentrations are [N₂]=0.038 M, [H₂]=0.012 M, [H₂O]=0.176 M, and the equilibrium constant Kc is approximately 533.

Step-by-step solution

Write the balanced equilibrium equation and the expression for Kc

The balanced equilibrium equation is already given in the exercise statement: \[ 2 \mathrm{NO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) \] The equilibrium constant, Kc, can be expressed as: \[ K_{\mathrm{c}} = \frac{[\mathrm{N}_{2}][\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{NO}]^2[\mathrm{H}_{2}]^2} \]

Create an ICE table

In order to find the equilibrium concentrations of each substance involved, we will create an ICE (Initial, Change, Equilibrium) table. Note that all initial amounts are in moles and we need to convert them into molar concentrations by dividing them by the volume of the vessel (1.0 L). Initial concentrations: [NO] = 0.10 mol/L [H₂] = 0.050 mol/L [H₂O] = 0.10 mol/L [N₂] = 0 (as it's not initially present) Table: NO H₂ N₂ H₂O Initial 0.10 0.050 0 0.10 Change -x -x +x +2x Equilibrium 0.10-x 0.050-x x 0.10+2x At equilibrium, the exercise states [NO] = 0.062 M. We can use this information to solve for x.

Solve for x using the given equilibrium concentration of NO

Since [NO] = 0.062 M at equilibrium, we can set: \[ 0.10 - x = 0.062 \] Now, solve for x: \[ x = 0.10 - 0.062 = 0.038\]

Calculate the equilibrium concentrations of H₂, N₂, and H₂O

Using the x value, we can now find the equilibrium concentrations of H₂, N₂, and H₂O: [N₂] = x = 0.038 M [H₂] = 0.050 − x = 0.050 - 0.038 = 0.012 M [H₂O] = 0.10 + 2x = 0.10 + (2 × 0.038) = 0.176 M Therefore, at equilibrium: [N₂] = 0.038 M [H₂] = 0.012 M [H₂O] = 0.176 M

Calculate Kc using the equilibrium concentrations

We can now calculate Kc using the equilibrium concentrations we just found: \[ K_{\mathrm{c}} = \frac{[\mathrm{N}_{2}][\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{NO}]^2[\mathrm{H}_{2}]^2} = \frac{(0.038)(0.176)^2}{(0.062)^2(0.012)^2} \] Evaluate Kc: \[ K_{\mathrm{c}} \approx 533 \] So, the equilibrium constant Kc for this reaction at 300 K is approximately 533.

Key concepts

Equilibrium Constant (Kc)

The equilibrium constant, represented as \(K_{\mathrm{c}}\), is a crucial concept in chemical equilibrium. It is a number that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced chemical equation. This constant is unique for every reaction at a given temperature, and it helps us predict the direction of the reaction. If \(K_{\mathrm{c}}\) is much larger than 1, the reaction favors products; if it is much smaller than 1, it favors reactants.
In our example, the reaction is:

• \(2 \mathrm{NO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\)

The \(K_{\mathrm{c}}\) for this equation is calculated using the concentrations at equilibrium:\[ K_{\mathrm{c}} = \frac{[\mathrm{N}_{2}][\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{NO}]^2[\mathrm{H}_{2}]^2} \]

ICE Table

An ICE table is a helpful tool in chemistry that assists in organizing the calculations needed to determine concentrations at equilibrium. The acronym stands for Initial, Change, and Equilibrium. This method helps to visualize how the concentrations of reactants and products change as the reaction progresses to equilibrium.
Using an ICE table, we start with the initial molar concentrations of reactants and products. In our problem:

• \([\mathrm{NO}] = 0.10\, \text{mol/L}\)
• \([\mathrm{H}_{2}] = 0.050\, \text{mol/L}\)
• \([\mathrm{H}_{2}\mathrm{O}] = 0.10\, \text{mol/L}\)
• \([\mathrm{N}_{2}] = 0\) (not initially present)

During the reaction, these values change by \(x\) according to stoichiometry:

• \(\text{Change: } -x, -x, +x, +2x\)

The equilibrium row then shows the concentrations:

• \(\text{Equilibrium: } 0.10-x, 0.050-x, x, 0.10+2x\)

Molar Concentration

Molar concentration, often called molarity, is a way of expressing the concentration of a solute in a solution. It is defined as the amount of solute (in moles) divided by the volume of the solution (in liters). This is crucial for understanding reactions in solutions because it standardizes concentration and allows for direct comparison between reactive species.
For example, if we have \(0.10\, \text{mol}\) of \(\mathrm{NO}\) in a \(1.0\, \text{L}\) vessel, the molar concentration is \(0.10\, \text{M}\). This simplicity makes molarity a popular choice for lab calculations. In equilibrium problems, molar concentrations are used to calculate changes and determine equilibrium states.
Using our equilibrium data:

• \([\mathrm{NO}]\) drops to \(0.062\, \text{M}\) at equilibrium.
• Adjustments in concentrations of other species (\([\mathrm{H}_{2}], [\mathrm{N}_{2}], [\mathrm{H}_{2}\mathrm{O}]\)) are made based on stoichiometry and these molar relationships.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry involves the calculations based on the relationships between the amounts of reactants and products in a chemical reaction. The coefficients in a balanced chemical equation tell us the ratios in which reactants combine and products form.
For the reaction \(2 \mathrm{NO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\), we can deduce:

• 2 moles of \(\mathrm{NO}\) react with 2 moles of \(\mathrm{H}_{2}\).
• This results in 1 mole of \(\mathrm{N}_{2}\) and 2 moles of \(\mathrm{H}_{2}\mathrm{O}\).

This tells us that every time \(x\) moles of \(\mathrm{NO}\) and \(\mathrm{H}_{2}\) decrease, \(x\) moles of \(\mathrm{N}_{2}\) form, and \(2x\) moles of \(\mathrm{H}_{2}\mathrm{O}\) form.
Understanding stoichiometry is essential for setting up an ICE table and correctly predicting changes as a reaction proceeds to equilibrium.