Problem 71 - Chemistry Matter and Change

Question

At \(350^{\circ} \mathrm{C}, K_{\mathrm{eq}}=1.67 \times 10^{-2}\) for the reversible reaction \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}^{2}(\mathrm{g})+\mathrm{I}^{2}(\mathrm{g}) .\) What is the concentration of HI at equilibrium if \(\left[\mathrm{H}^{2}\right]\) is \(2.44 \times 10^{-3} M\) and \(\left[\mathrm{I}^{2}\right]\) is \(7.18 \times 10^{-5} \mathrm{M}\) ?

Step-by-Step Solution

Answer

The concentration of \([\text{HI}]\) at equilibrium is approximately \(3.24 \times 10^{-3} \text{ M}\).

Step 1: Write the Expression for the Equilibrium Constant

For the reaction \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}^{2}(\mathrm{g})+\mathrm{I}^{2}(\mathrm{g})\), the equilibrium constant expression is given by \(K_{eq} = \frac{[\mathrm{H}_2][\mathrm{I}_2]}{[\mathrm{HI}]^2}\). Given that \(K_{eq} = 1.67 \times 10^{-2}\).

Step 2: Plug in the Known Values

Substitute the known values \([\mathrm{H}_2] = 2.44 \times 10^{-3} \text{ M}\) and \([\mathrm{I}_2] = 7.18 \times 10^{-5} \text{ M}\) into the expression: \[1.67 \times 10^{-2} = \frac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{[\mathrm{HI}]^2}\]

Step 3: Solve for \([\mathrm{HI}]^2\)

Compute the numerator: \[(2.44 \times 10^{-3})(7.18 \times 10^{-5}) = 1.75 \times 10^{-7}\] Then, solve for \([\mathrm{HI}]^2\):\[[\mathrm{HI}]^2 = \frac{1.75 \times 10^{-7}}{1.67 \times 10^{-2}}\] This gives: \[[\mathrm{HI}]^2 = 1.047 \times 10^{-5}\]

Step 4: Solve for \([\mathrm{HI}]\)

Take the square root of both sides to find \([\mathrm{HI}]\):\[[\mathrm{HI}] = \sqrt{1.047 \times 10^{-5}}\]Evaluating this gives:\([\mathrm{HI}] \approx 3.24 \times 10^{-3} \text{ M}\).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_{eq}\), is a crucial part of understanding chemical equilibrium. It is a numerical value that tells us the ratio of the concentrations of the products and reactants for a particular reversible reaction at equilibrium. This value is specific for each reaction and varies with temperature. In our exercise, \(K_{eq} = 1.67 \times 10^{-2}\) at \(350^{\circ} \text{C}\) for the reaction \(2 \text{HI(g)} \rightleftharpoons \text{H}_2 \text{(g)} + \text{I}_2 \text{(g)}\). This means that at this specified temperature, the ratio comes from the concentration of hydrogen and iodine gases over the concentration of hydrogen iodide raised to its stoichiometric coefficient. The equilibrium constant helps predict which direction the reaction will favor—this way we can understand whether products or reactants are more prevalent at equilibrium.

Concentration

Concentration refers to the amount of a substance in a given volume. It's often expressed in molarity (M), which is moles per liter. In chemical reactions, concentration plays a significant role in determining how the reaction proceeds towards equilibrium. In the given exercise, concentrations are provided for the products \([\text{H}_2] = 2.44 \times 10^{-3} \text{ M}\) and \([\text{I}_2] = 7.18 \times 10^{-5} \text{ M} \). These values are essential to calculate the concentration of the reactant \([\text{HI}]\) at equilibrium by using the equilibrium constant equation. The concentration of substances helps us to understand the dynamics and extent of the reaction, particularly how much of each species is present when the system is at equilibrium.

HI Equilibrium

The concept of HI (hydrogen iodide) equilibrium revolves around reaching a balanced state in the reaction system \(2\text{HI(g)} \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)\). At equilibrium, the concentration of \(\text{HI}\) stops changing because the rate of its formation and decomposition become equal. This scenario is what we solve for in the exercise, aiming to find the equilibrium concentration of \(\text{HI}\) using known values for \([\text{H}_2]\) and \([\text{I}_2]\) as well as the \(K_{eq}\). In essence, equilibrium for this reaction means that even though \(\text{HI}\), \(\text{H}_2\), and \(\text{I}_2\) are constantly reacting and forming each other, their concentrations remain constant over time.

Reversible Reaction

A reversible reaction is one where the reactants form products, which can in turn react together to form the reactants. For the equation \(2 \text{HI(g)} \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)\), this reversibility is represented by the double-headed arrow. Unlike irreversible reactions, reversible reactions can reach a state of equilibrium where the forward and reverse chemical reactions occur at the same rate, resulting in no net change in concentrations of the reactants and products over time. Understanding the reversible nature of reactions is key to predicting how a system can be manipulated. For instance, if the concentration of one component is changed, the system will shift to re-establish equilibrium. This concept is also described by Le Chatelier's principle. Reversible reactions are foundational in chemical processes because they allow control over the production yield and the refinement of chemical substances.