Question

Gaseous hydrogen iodide is placed in a closed container at \(450^{\circ} \mathrm{C},\) where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) At equilibrium it is found that \([\mathrm{HI}]=4.50 \times 10^{3} \mathrm{M},\left[\mathrm{H}_{2}\right]=5.75 \times 10^{4} \mathrm{M}\), and \(\left[\mathrm{I}_{2}\right]=5.75 \times 10^{-4} \mathrm{M}\). What is the value of \(K_{c}\) at this temperature?

Short answer

The value of the equilibrium constant (\(K_c\)) for the decomposition of hydrogen iodide at \(450^{\circ} \mathrm{C}\) is \(1.632 \times 10^{-7}\).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the decomposition of hydrogen iodide is: \[2\,\mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g)\]

Write the expression for the equilibrium constant

The equilibrium constant expression for the given reaction is given by: \[K_c = \frac{[\mathrm{H}_2][\mathrm{I}_2]}{[\mathrm{HI}]^2}\]

Substitute the given equilibrium concentrations

The given equilibrium concentrations are: \[[\mathrm{HI}] = 4.50 \times 10^3 \mathrm{M}\] \[[\mathrm{H}_2] = 5.75 \times 10^4 \mathrm{M}\] \[[\mathrm{I}_2] = 5.75 \times 10^{-4} \mathrm{M}\] Substitute these values into the equilibrium constant expression: \[K_c = \frac{(5.75 \times 10^4 \mathrm{M})(5.75 \times 10^{-4} \mathrm{M})}{(4.50 \times 10^3 \mathrm{M})^2}\]

Calculate \(K_c\)

Now calculate the value of \(K_c\) by performing the arithmetic operations: \[K_c = \frac{(5.75 \times 10^4)(5.75 \times 10^{-4})}{(4.50 \times 10^3)^2} = \frac{3.30625 \times 10^{0}}{2.025 \times 10^7}\] \[K_c = 1.632 \times 10^{-7}\] So, the value of the equilibrium constant (\(K_c\)) for the decomposition of hydrogen iodide at \(450^{\circ} \mathrm{C}\) is \(1.632 \times 10^{-7}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating state in a chemical reaction where the concentrations of reactants and products stop changing. This does not mean the reactions halt, but instead, the forward and reverse reactions occur at equal rates. At this point, the system reaches a stable condition known as dynamic equilibrium. This state is crucial because it determines how far the reaction proceeds and is especially important in predicting the concentration of products and reactants at any given time. When dealing with gaseous hydrogen iodide decomposing into hydrogen and iodine, we see equilibrium being achieved after a while when concentrations stabilize. It is important to recognize that the conditions, such as temperature and pressure, significantly influence equilibrium. For the case at hand, this equilibrium happens at a high temperature of 450°C.

Reaction Kinetics

Reaction kinetics involve exploring how fast a chemical reaction proceeds. When looking at the decomposition of hydrogen iodide into hydrogen and iodine, reaction kinetics helps us understand the rate at which these products form. It focuses on factors such as the concentration of reactants, temperature, and catalysts which might affect the speed of the chemical process. In our exercise, we consider the temperatures and specific concentrations, contributing to the reaction rate and leading the system toward equilibrium. Understanding kinetics can determine how quickly equilibrium is reached and how the concentrations alter before stabilizing. This knowledge helps in controlling practical applications like industrial processes, ensuring they run efficiently and effectively.

Concentration Calculations

Concentration calculations are key when analyzing chemical equilibria in a reaction. The task involves determining how much of each substance is present in a mixture, crucial for deriving the equilibrium constant, denoted as \(K_c\). To find \(K_c\), which is a constant that provides insight into the position of equilibrium, you must measure the concentration of the products and reactants at equilibrium. Considering the balanced equation, \(K_c\) is calculated using the formula:

• The concentrations of the products are multiplied in the numerator.
• The concentration of the reactants, raised to their stoichiometric coefficients, is in the denominator.

Using this structured approach, the values from the exercise help us determine \(K_c\) at 450°C. For our example, after substituting the provided concentrations into the expression, arithmetic operations yield the equilibrium constant as \(1.632 \times 10^{-7}\). This value explains the equilibrium position of the reaction and helps predict how the reaction will behave under similar conditions.