Problem 65 - Chemistry: Principles and Reactions

Question

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

Step-by-Step Solution

Answer

Answer: 95.2%

Step 1: Determine the initial moles of isopropyl alcohol

Given, 20.0 mL of isopropyl alcohol with a density of 0.785 g/mL. First, we convert the volume to mass using the density: $$ \text{Mass of isopropyl alcohol} = \text{Volume} \times \text{Density} = 20.0 \,\text{mL} \times 0.785 \, \frac{\text{g}}{\text{mL}} = 15.7\,\text{g} $$ Now, we need to convert this mass to moles using the molar mass of isopropyl alcohol (Molar mass: \(C_3H_7OH = 3(12.01) + 7(1.01) + 16.00 + 1.01 = 60.11 \,\text{g/mol}\)): $$ \text{Moles of isopropyl alcohol} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{15.7\,\mathrm{g}}{60.11\,\frac{\mathrm{g}}{\mathrm{mol}}} = 0.261\,\mathrm{mol} $$

Step 2: Set up the equilibrium expression

At equilibrium, the ratio of the products to the reactants is constant and equal to the equilibrium constant (K). The reaction considered is: $$ \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ Let x be the amount of isopropyl alcohol that decomposes at equilibrium. Then at equilibrium, we will have: - Moles of isopropyl alcohol = \(0.261 - x\) mol - Moles of acetone = x mol - Moles of hydrogen gas = x mol The equilibrium expression is given by: $$ K = \frac{[\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)][\mathrm{H}_{2}(g)]}{[\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g)]} $$

Step 3: Calculate the equilibrium concentrations

Since the volume of the vessel is 5.00 L, we can calculate the concentrations of each species at equilibrium: - Isopropyl alcohol: \(\frac{0.261 - x}{5.00} = 0.0522 - 0.20x\) - Acetone: \(\frac{x}{5.00} = 0.20x\) - Hydrogen gas: \(\frac{x}{5.00} = 0.20x\) Now plug these concentrations into the equilibrium expression and solve for x: $$ 0.45 = \frac{(0.20x)(0.20x)}{0.0522 - 0.20x} $$

Step 4: Solve for x

Solve the above equation for x: $$ x = 0.0124\,\mathrm{mol} $$

Step 5: Calculate the percentage of isopropyl alcohol remaining undissociated

Now we can find the percentage of isopropyl alcohol that remains undissociated: $$ \text{Percentage undissociated} = \frac{0.261 - 0.0124}{0.261} \times 100 = 95.2\% $$ Thus, 95.2% of the isopropyl alcohol remains undissociated at equilibrium.

Key Concepts

Isopropyl Alcohol Decomposition

Understanding the decomposition of isopropyl alcohol is essential when studying chemical reactions and equilibrium. Isopropyl alcohol, commonly used as a disinfectant, has a tendency to break down into acetone and hydrogen gas, particularly when exposed to higher temperatures like 180°C. This process is reversible and can achieve a state where the rate of decomposition equals the rate of formation—a dynamic condition known as chemical equilibrium.

During this reaction, the original substance, isopropyl alcohol, is known as the reactant, while the resultant substances—acetone and hydrogen gas—are called the products. The understanding of this decomposition helps in many industrial processes, where controlling the amount of each substance is crucial for the yield and quality of the final products.

Visualizing the Decomposition

Imagine a scenario where isopropyl alcohol molecules are constantly colliding and breaking down into acetone and hydrogen. Simultaneously, acetone and hydrogen molecules occasionally collide and reform the original alcohol. When these two rates balance out, the system has reached equilibrium.

Equilibrium Constant

The equilibrium constant (K) is a number that provides a measure of the extent to which a reaction proceeds to products at a given temperature. It is unique for every chemical reaction. The larger the value of K, the more the reaction favors the formation of products at equilibrium. Conversely, a smaller K value indicates that the reactants are favored.

This constant is derived from the concentrations of the reactants and products at equilibrium. Mathematically, for the decomposition of isopropyl alcohol, K is defined as the concentration of acetone times the concentration of hydrogen gas, divided by the concentration of isopropyl alcohol. The equilibrium constant is dimensionless as it is a ratio of concentrations, each divided by their standard state concentration, normally 1 M.

Finding Equilibrium

If the value of K is known, like the 0.45 in our exercise, scientists can use it to predict the direction of the reaction or to calculate unknown concentrations of reactants and products at equilibrium.

Mole Calculation

In chemistry, the mole is a fundamental unit in the Système International d'Unités (SI) which is used for measuring the amount of substance. One mole corresponds to the number of atoms found in 12 grams of carbon-12, which is approximately 6.022 x 10^23 atoms, known as Avogadro's number.

Mole calculations are vital for understanding chemical reactions, as they allow us to count particles of substances by weighing them. With the molar mass—a quantity that gives the mass of 1 mole of a substance—we can convert between the mass of a substance and the number of moles it contains. In the exercise, this is done by taking the mass of isopropyl alcohol and dividing it by its molar mass to find the initial moles present.

Relating Moles to Real-World Problems

Once the number of moles of each substance in a reaction is known, you can perform a variety of calculations, including determining the percentage of a reactant that remains undissociated in equilibrium, as was required in the given exercise. Additionally, understanding the mole concept helps in the accurate measurement of chemicals for reactions, ensuring safety and efficiency in chemical experimentation and production.