Problem 100 - Chemistry An Atoms First Approach

Question

A \(4.72-\mathrm{g}\) sample of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) was placed in an otherwise empty 1.00 -L flask and heated to \(250 .^{\circ} \mathrm{C}\) to vaporize the methanol. Over time, the methanol vapor decomposed by the \(10=\) following reaction:$$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$.After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much \(\mathrm{H}_{2}(g)\) as \(\mathrm{CH}_{3} \mathrm{OH}(g) .\) Calculate \(K\) for this reaction at \(250 .^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The equilibrium constant K for the decomposition reaction of methanol (CH3OH) at 250°C is approximately 54.98.

Step 1: Find the initial moles of methanol

We can find the initial moles of methanol using the given mass (4.72 g) and the molar mass of methanol: Initial moles of CH3OH = (4.72 g) / (32.04 g/mol) = 0.147 mol

Step 2: Determine the moles of reactants and products at equilibrium

Let x be the amount of CH3OH decomposed at equilibrium. Given that the ratio of H2 to CH3OH at equilibrium is 33:1, then the amount of H2 produced at equilibrium would be 33x. Equilibrium moles of CH3OH = (0.147 - x) mol Equilibrium moles of CO = x mol Equilibrium moles of H2 = 33x mol Note that the initial moles of CO and H2 are 0.

Step 3: Write the equilibrium expression for K

The equilibrium expression for the given reaction is: K = [CO][H2]^2 / [CH3OH]

Step 4: Calculate K using equilibrium concentrations

First, we need to find the concentrations of each species at equilibrium. Since the volume of the flask is 1.00 L, equilibrium concentrations can be represented by the moles of each species: [CH3OH] = (0.147 - x) mol/L [CO] = x mol/L [H2] = 33x mol/L Now, we need to find the value of x. We know the ratio of H2 to CH3OH at equilibrium is 33:1, which means: (H2) / (CH3OH) = (33x) / (0.147 - x) = 33 Solving for x, we find x = 0.144 Now, we can plug the values of x into our equilibrium concentrations: [CH3OH] = (0.147 - 0.144) mol/L = 0.003 mol/L [CO] = 0.144 mol/L [H2] = 33 * 0.144 mol/L = 4.752 mol/L Finally, we can plug these concentrations into the K expression to find K: K = (0.144)(4.752)^2 / (0.003) = 54.98 The equilibrium constant K for this reaction at 250°C is approximately 54.98.

Key Concepts

Molar Mass Calculation

Understanding the concept of molar mass is critical when studying chemical reactions. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's the sum of the atomic masses of all the atoms in a molecular formula. For methanol (\textbf{CH}\(_3\)\textbf{OH}), it consists of one carbon atom, four hydrogen atoms, and one oxygen atom. To calculate the molar mass, you would add the atomic masses of these atoms according to the periodic table:

\textbf{Carbon (C):} 12.01 g/mol
\textbf{Hydrogen (H):} 1.01 g/mol x 4 = 4.04 g/mol
\textbf{Oxygen (O):} 16.00 g/mol

When you sum these values, you get the molar mass of methanol: 12.01 + 4.04 + 16.00 = 32.05 g/mol. With this information, you can then calculate how many moles are in a given mass of the substance by dividing the mass by the molar mass, as seen in the exercise solution where the initial moles of methanol were determined.

Equilibrium Expression

The equilibrium expression, represented by the symbol K, quantifies the relative concentrations of reactants and products at equilibrium. For the reaction \textbf{CH}\(_3\)\textbf{OH}(g) \(\rightleftharpoons\) CO(g) \(+\) 2 H\(_2\)(g), the equilibrium expression is written as:K = [CO][H\(_2\)]\(_2\) / [CH\(_3\)OH]Here, the square brackets represent the concentration of each species in moles per liter (mol/L), and the exponents represent the coefficients from the balanced equation. In equilibrium calculations, it's important to note that only the gaseous and aqueous species are included in the expression, while solids and pure liquids are excluded. For this methanol decomposition reaction, the concentration of each species is used to solve for K, which tells us how far the reaction proceeds towards products at a certain temperature.

Effusion of Gases

Effusion is the process by which gas molecules pass through a tiny opening into a vacuum. According to Graham's law of effusion, lighter gas molecules effuse faster than heavier ones. This can be used to compare the rates of effusion for different gases. In the given exercise, this concept is applied to examine the composition of the effusing gas. The measurements indicated that hydrogen effused more rapidly, which makes sense as hydrogen (H\(_2\)) has the lowest molar mass among common gaseous substances. The 33:1 ratio of H\(_2\) to CH\(_3\)OH in the effusing gas helped to calculate the equilibrium concentrations for the reaction.

Mole Concept

The mole concept is a fundamental principle in chemistry that provides a bridge between the atomic scale and the macroscopic world. One mole, approximately \(6.022 \times 10^{23}\) entities (Avogadro's number), is the amount of substance that contains as many particles as there are atoms in exactly 12 grams of carbon-12. This concept allows chemists to count atoms and molecules by weighing them. In the context of the reaction in the exercise, understanding the mole concept is critical for determining the amounts of reactants and products. The balance between the decomposed methanol and the produced carbon monoxide and hydrogen followed the mole concept, which also guided the successful calculation of the equilibrium constant (K) for the methanol decomposition reaction.