Question

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C}\), where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\). At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\), and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\). What is the value of \(K_{c}\) at this temperature?

Short answer

At a temperature of \(425^{\circ} \mathrm{C}\), the equilibrium constant Kc for the given reaction is approximately \(0.0184\).

Step-by-step solution

Write the equilibrium constant expression

The balanced equation for the reaction is given: \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g})\). From this equation, we will write the equilibrium constant expression: \(K_c = \frac{[\mathrm{H}_{2}] [\mathrm{I}_{2}]}{[\mathrm{HI}]^2}\)

Plug in the given equilibrium concentrations

We are given the equilibrium concentrations: \([\mathrm{HI}] = 3.53 \times 10^{-3} \mathrm{M}\) \([\mathrm{H}_{2}] = 4.79 \times 10^{-4} \mathrm{M}\) \([\mathrm{I}_{2}] = 4.79 \times 10^{-4} \mathrm{M}\) Substitute these values into the equilibrium constant expression: \(K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2}\)

Evaluate the expression to find Kc

Now, we'll calculate Kc: \(K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2} \approx \frac{2.29841 \times 10^{-7}}{1.24549 \times 10^{-5}} \approx 0.0184\) At a temperature of \(425^{\circ} \mathrm{C}\), the equilibrium constant Kc is approximately 0.0184.

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. As a result, the concentrations of the reactants and products remain constant over time, not because the reactions have stopped, but because they are occurring simultaneously at an equal pace. In your textbook problem, the decomposition of hydrogen iodide gas into hydrogen and iodine gas is an example of a reversible reaction that has reached equilibrium at a given temperature.

To visualize this, consider two children on a seesaw who have found a balance. Neither side of the seesaw moves up or down, similar to how the concentrations of reactants and products in a chemical system at equilibrium don't change. The key to understanding equilibrium lies in knowing that it is a dynamic state where reactions still occur, but no net change in concentration can be observed.

Reaction Quotient

The reaction quotient, denoted as Q, is a measure identical in form to the equilibrium constant but for any set of conditions, not just at equilibrium. It is used to predict the direction of the chemical reaction. To calculate Q, you use the same expression as the equilibrium constant (Kc), except the concentrations used are those at a moment in time, not necessarily at equilibrium.

For example, if you start with certain concentrations of reactants and products and Q is less than Kc, the reaction will proceed forwards (towards products) to reach equilibrium. Conversely, if Q is greater than Kc, the reaction will proceed in the reverse direction (towards reactants) to reach equilibrium. When Q equals Kc, the system is already at equilibrium, signifying no net change in the concentrations of reactants and products.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of reactants and products in a chemical reaction that are present when the reaction has reached equilibrium. These concentrations can be plugged into the equilibrium expression to calculate the equilibrium constant, Kc, which is a unique value for each reaction at a given temperature.

The calculations, as shown in the exercise solution, involve substituting these concentrations into the equilibrium expression correctly following stoichiometry. A crucial point to remember is that the concentrations used are those at equilibrium where no further change is detected, providing a snapshot of the system's balanced state.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a system at equilibrium responds to changes in concentration, temperature, or pressure. According to this principle, if an external stress is applied to a system at equilibrium, the system adjusts to minimize the stress and restore a new equilibrium state.

For instance, if more reactant is added to the system, Le Chatelier's principle predicts that the equilibrium will shift towards forming more product to relieve this stress. Conversely, if the temperature is increased for an exothermic reaction, the system will shift towards the reactants to absorb the excess heat. Understanding this principle is essential for manipulating and controlling chemical reactions, especially in industrial chemical processes.