Problem 34 - Chemistry The Central Science

Question

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C},\) where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) 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?

Step-by-Step Solution

Answer

The equilibrium constant, \(K_c\), for the reaction \(2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)\) at the given temperature and concentrations is approximately 0.035.

Step 1: 1. Write down the balanced chemical equation

First, we write the balanced chemical equation for the given reaction: \[ 2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g) \]

Step 2: 2. Write the expression for the equilibrium constant, \(K_c\)

For the given balanced chemical equation, the equilibrium constant expression is: \[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]

Step 3: 3. Substitute the given equilibrium concentrations into the \(K_c\) expression

We are given the equilibrium concentrations of each species: \([HI] = 3.53 \times 10^{-3} M\), \([H_2] = 4.79 \times 10^{-4} M\), and \([I_2] = 4.79 \times 10^{-4} M\). Substitute these values into the \(K_c\) expression: \[ K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2} \]

Step 4: 4. Calculate \(K_c\)

Now, simply perform the calculation: \[ K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2} \approx 0.035 \]

Step 5: 5. State the final result

The equilibrium constant, \(K_c\), for the given reaction and temperature is approximately 0.035.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\), is a crucial concept in understanding how reactions reach a balance between the products and reactants. It is a measure of the ratio of concentrations of products to reactants, each raised to the power of their respective coefficients from the balanced equation.
For example, in the decomposition reaction of hydrogen iodide, \(2HI(g)\), into hydrogen \(H_2(g)\) and iodine \(I_2(g)\), the equilibrium constant expression is given as: \[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
The concentrations of the products \([H_2]\) and \([I_2]\) are multiplied in the numerator, while the concentration of the reactant \([HI]\) is squared in the denominator, based on its coefficient in the balanced equation.

• If \(K_c\) is a large number, it indicates that the reaction favors products at equilibrium.
• If \(K_c\) is a small number, the reaction favors reactants at equilibrium.

Understanding \(K_c\) helps chemists predict the direction of the reaction and its degree of completion at a given temperature.

Decomposition Reaction

A decomposition reaction involves a single compound breaking down into two or more simpler substances. These reactions are important in fields such as chemistry and industry because they often release energy and form essential materials.
In our given exercise, hydrogen iodide undergoes decomposition: \[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]
Here, the reaction is reversible, indicated by the double arrow, meaning both the forward and reverse reactions occur until equilibrium is reached.
Some key points about decomposition reactions include:

• They can be triggered by heat, light, or electricity.
• Often involve a single reactant breaking down into multiple products.
• Have practical applications, such as in the decomposition of material for recycling or energy generation.

In a closed system, like the one in our exercise, the decomposition continues until a point where the rate of forming products equals the rate of reforming reactants, leading to equilibrium.

Concentration Calculation

Calculating concentrations is a vital skill in chemistry that allows us to quantify how much of each reactant or product is present in a chemical reaction at any given time. Concentration is usually expressed in moles per liter (M), also known as molarity.
In this exercise, we calculate concentrations to evaluate the equilibrium constant \(K_c\). Given values are:

• \([HI] = 3.53 \times 10^{-3} \text{ M}\)
• \([H_2] = 4.79 \times 10^{-4} \text{ M}\)
• \([I_2] = 4.79 \times 10^{-4} \text{ M}\)

These concentrations are substituted into the \(K_c\) expression: \[ K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2} \]
This formula works because it correlates the concentrations of the reactants and products at equilibrium, allowing us to calculate \(K_c\) accurately. Online tools or calculators often assist with these calculations, but understanding the underlying principles is essential for doing it manually and grasping the concept's significance.