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 \(5.00-\mathrm{L}\) vessel and heated to \(180^{\circ} \mathrm{C},\) what percentage remains undissociated at equilibrium?

Short answer

Answer: Approximately 16.5% of isopropyl alcohol remains undissociated at equilibrium.

Step-by-step solution

Determine the initial moles of isopropyl alcohol

Given the initial volume and density of isopropyl alcohol, we can find the mass by multiplying volume and density. Then, we can convert the mass to moles by dividing by its molar mass. Initial volume = 20.0 mL Density = 0.785 g/mL Molar mass of isopropyl alcohol (C3H7OH) = 3×12.01 + 7×1.01 + 1×16.00 = 60.11 g/mol First, find the mass: Mass = Initial volume × Density = 20.0 mL × 0.785 g/mL = 15.7 g Now, convert mass to moles: Initial moles of isopropyl alcohol = Mass / Molar mass = 15.7 g / 60.11 g/mol ≈ 0.261 moles

Set up the equilibrium expression

To set up the equilibrium expression, let's use x to represent the moles of isopropyl alcohol that decompose. Then, at equilibrium, we will have: - Moles of isopropyl alcohol (C3H7OH): 0.261 - x - Moles of acetone (C2H6CO): x - Moles of hydrogen gas (H2): x The equilibrium constant, K, is given as 0.45. We can set up the equilibrium expression as follows: K = [C2H6CO][H2] / [C3H7OH]

Solve for x

From the equilibrium expression, substitute the moles of each species: 0.45 = (x)(x) / (0.261 - x) Now, solve the equation for x: 0.45(0.261 - x) = x^2 0.11745 - 0.45x = x^2 x^2 + 0.45x - 0.11745 = 0 Now, we have a quadratic equation. Since the amount of decomposed isopropyl alcohol cannot be negative, we will consider only the positive root. x ≈ 0.218 moles

Calculate the percentage remaining undissociated

Now that we have the moles of isopropyl alcohol that decompose, we can calculate the moles remaining undissociated and the percentage. Moles remaining undissociated = Initial moles - Decomposed moles = 0.261 - 0.218 ≈ 0.043 moles Percentage remaining undissociated = (Moles remaining undissociated / Initial moles) × 100 = (0.043 / 0.261) × 100 ≈ 16.5% Therefore, at equilibrium, approximately 16.5% of isopropyl alcohol remains undissociated.