Problem 90 - General Chemistry: Principles and Modern Applications

Question

At \(500 \mathrm{K}\), a 10.0 L equilibrium mixture contains 0.424 \(\mathrm{mol} \mathrm{N}_{2}, 1.272 \mathrm{mol} \mathrm{H}_{2},\) and \(1.152 \mathrm{mol} \mathrm{NH}_{3} .\) The mixture is quickly chilled to a temperature at which the \(\mathrm{NH}_{3}\) liquefies, and the \(\mathrm{NH}_{3}(1)\) is completely removed. The 10.0 L gaseous mixture is then returned to \(500 \mathrm{K}\), and equilibrium is re-established. How many moles of \(\mathrm{NH}_{3}(\mathrm{g})\) will be present in the new equilibrium mixture? $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3} \quad K_{\mathrm{c}}=152 \text { at } 500 \mathrm{K}$$

Step-by-Step Solution

Answer

The number of moles of \(\mathrm{NH}_3\) at the new equilibrium is obtained from solving the equilibrium constant expression. It is an algebraic problem that can be solved by standard methods.

Step 1: Expression of reaction quotient and writing the expression for the equilibrium constant

The reaction to consider is \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}\). Due to the chilling process, all \(\mathrm{NH}_{3}\) is removed initially. Therefore, when equilibrium is re-established, let x be the number of moles of \(\mathrm{NH}_{3}\) formed. This means 0.5x mol of \(\mathrm{N}_{2}\) and 1.5x mol of \(\mathrm{H}_{2}\) will react to produce x mol of \(\mathrm{NH}_{3}\), given 1 mol of \(\mathrm{N}_{2}\) reacts with 3 mol of \(\mathrm{H}_{2}\) to produce 2 mol of \(\mathrm{NH}_{3}\). The expression for the equilibrium constant \(K_{c}\) is therefore \([NH_{3}]^{2}/([N_{2}][H_{2}]^{3})\), where [A] stands for molarity of A.

Step 2: Calculation of molarities

Before reaction, the molarities of \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) are 0.424 mol/10.0 L, 1.272 mol/10.0 L, and 0 mol/10.0 L respectively. After reaction, the molarities become (0.424 - 0.5x) mol/10.0 L, (1.272 - 1.5x) mol/10.0 L, and x mol/10.0 L respectively.

Step 3: Substitution and solving for x

Substituting the expressions for molarities into the expression for \(K_{c}\) and equating it to the given value of 152, the equation to solve for x becomes \((x/10.0)^{2}/(((0.424 - 0.5x)/10.0)((1.272 - 1.5x)/10.0)^{3}) = 152\). Solving for x yields the number of moles of \(\mathrm{NH}_{3}\) at the new equilibrium.

Key Concepts

Reaction Quotient

The reaction quotient, denoted as Q, plays a pivotal role in predicting the direction in which a chemical reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant (K), but with the initial concentrations or partial pressures of the reactants and products instead of the concentrations at equilibrium.

For the reaction \( \mathrm{N_2}(g) + 3 \mathrm{H_2}(g) \rightleftharpoons 2 \mathrm{NH_3}(g) \), the reaction quotient is given by \( Q = \frac{[NH_3]^2}{[N_2][H_2]^3} \), where the brackets denote the molarity of the compounds at a specific time before equilibrium is reached. If Q is less than the equilibrium constant, K, the reaction will proceed in the forward direction to form more products. Conversely, if Q is greater than K, the reaction will shift to the left, forming more reactants until equilibrium is achieved.

Equilibrium Constant

The equilibrium constant, K, reflects the ratio of product concentrations to reactant concentrations at equilibrium. For a balanced chemical equation \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).

In the provided exercise, the equilibrium constant expression for the synthesis of ammonia is \( K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \). The subscript 'c' denotes that the concentrations are measured in molarity. The value of the equilibrium constant \( K_c = 152 \), indicating the relative amounts of each substance present when the system is in a state of equilibrium at 500 K. The equilibrium constant is a fundamental parameter in predicting whether a reaction mixture will proceed to form more products or reactants.

Molarity

Molarity is a measure of concentration used in chemistry to express the amount of a substance in a given volume of solution. It is defined as the number of moles of solute per liter of solution (mol/L).

In the exercise, the initial molarities of the reactants \( \mathrm{N_2} \) and \( \mathrm{H_2} \) are calculated by dividing their mole amounts by the volume of the container, which is 10.0 L. For instance, the molarity of \( \mathrm{N_2} \) is calculated as \( 0.424 \, \text{mol} / 10.0 \, \text{L} \). Molarity plays an essential role in determining reaction quotient and equilibrium expressions since these calculations require concentrations of reactants and products.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a system at equilibrium responds to external changes. It states that if an equilibrium system is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract the imposed change and a new equilibrium will be established.

In the given problem, when the \( \mathrm{NH_3} \) is removed from the system, the equilibrium is disrupted. According to Le Chatelier's principle, the reaction will shift to produce more \( \mathrm{NH_3} \) to restore equilibrium. Understanding this principle helps to predict the effect of removing \( \mathrm{NH_3} \) from the mixture, and permits the calculation of the new mole amounts of \( \mathrm{NH_3} \) that will be present once the system reaches equilibrium again.