Problem 110 - Chemistry

Question

A 4.72 -g sample of methanol (CH_ 3 \(\mathrm{OH}\) ) was placed in an otherwise empty \(1.00-\mathrm{L}\) flask and heated to \(250 .^{\circ} \mathrm{C}\) to vaporize the methanol. Over time, the methanol vapor decomposed by the 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 \(250 . \mathrm{C} .\)

Step-by-Step Solution

Answer

The equilibrium constant for the decomposition of methanol vapor at \(250\,\mathrm{C}\) is \(K \approx 7.94 \times 10^{-4}\).

Step 1: Calculate initial moles of methanol

First, we need to find the initial moles of methanol that was placed in the flask. We can use the mass of methanol given and its molar mass to determine this. The molar mass of methanol, \(\mathrm{CH}_3\mathrm{OH}\), is 12.01 (C) + 1.008 (H) + 1.008 (H) + 1.008 (H) + 16.00 (O) + 1.008 (H) = 32.042 g/mol. Now, we can find the moles of methanol: moles of \(\mathrm{CH}_3\mathrm{OH} = \frac{4.72 \text{g}}{32.042 \text{g/mol}} \approx 0.147\text{ moles}\).

Step 2: Set up ICE table

To find the equilibrium concentrations, we will use the ICE (Initial, Change, Equilibrium) analysis: Initial: \(\mathrm{CH}_3\mathrm{OH}(g)\): \(0.147 \mathrm{mol/L}\) \(\mathrm{CO}(g) + 2 \mathrm{H}_2(g)\): \(0 \mathrm{mol/L}\) Change: \(\mathrm{CH}_3\mathrm{OH}(g)\): -\(x \mathrm{mol/L}\) \(\mathrm{CO}(g) + 2 \mathrm{H}_2(g)\): +\(x\) \(\mathrm{mol/L}, + 2x \,\mathrm{mol/L}\) Equilibrium: \(\mathrm{CH}_3\mathrm{OH}(g)\): \((0.147 - x)\,\mathrm{mol/L}\) \(\mathrm{CO}(g)\): \(x\,\mathrm{mol/L}\) \(\mathrm{H}_2(g)\): \(2x\,\mathrm{mol/L}\)

Step 3: Determine equilibrium concentrations using given ratio

We're given that the ratio of \(\mathrm{H}_{2}(g)\) to \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is 33.0 in the effusing gas. Using the equilibrium concentrations, we can write this equation: \(\frac{2x}{x} = \frac{\mathrm{H}_2}{\mathrm{CH}_3\mathrm{OH}} = 33\) Solving for \(x\), we get \(x = \frac{2}{33}\). Since \(x < 0.147\), this small initial approximation is valid. Now, we can find the equilibrium concentrations: \(\mathrm{CH}_3\mathrm{OH} = 0.147 - x \approx 0.141\,\mathrm{mol/L}\) \(\mathrm{CO} = x \approx \frac{2}{33}\,\mathrm{mol/L}\) \(\mathrm{H}_2 = 2x\approx \frac{4}{33}\,\mathrm{mol/L}\)

Step 4: Find the equilibrium constant (K)

Now, we can use the calculated equilibrium concentrations to find the equilibrium constant: \(K = \frac{[\mathrm{CO}] [\mathrm{H}_2]^2}{[\mathrm{CH}_3\mathrm{OH}]} = \frac{\frac{2}{33} \times \left(\frac{4}{33}\right)^2}{0.141}\) Evaluating this expression, we get: \(K \approx 7.94 \times 10^{-4}\) So, at \(250\,\mathrm{C}\), the equilibrium constant for the decomposition of methanol vapor is \(K \approx 7.94 \times 10^{-4}\).

Key Concepts

ICE Table

The ICE table is a tool used in chemistry to represent the changes in concentrations of reactants and products as a chemical reaction reaches equilibrium. It's an acronym for **Initial, Change,** and **Equilibrium**, and involves setting up and solving equations based on these stages.

**Using an ICE Table:**
- **Initial concentrations:** Start by noting the initial concentration of each species involved in the reaction. For substances that do not appear initially, their concentration is set to zero.
- **Change:** Represent the changes that occur to each substance as the reaction progresses. If a reactant is consumed, its change value is negative, and if a product is formed, its change value is positive.
- **Equilibrium concentrations:** Calculate the final equilibrium concentrations using the initial values and the changes.

An ICE table can be particularly useful because it visually demonstrates how equilibrium is achieved. By writing out the concentrations step-by-step, it's easier to keep track of the stoichiometry and dynamic nature of chemical reactions.

Equilibrium Constant

The equilibrium constant, denoted as **K**, is a fundamental concept in chemical equilibrium. It is a number that expresses the ratio of the concentrations of products to reactants at equilibrium for a particular reaction, each raised to the power of their respective stoichiometric coefficients.

**Key Points about the Equilibrium Constant:**

• **Formula for K**: It can be written as \[K = \frac{[Products]}{[Reactants]}\]where the square brackets denote concentrations.
• **Constant Values**: The value of K is constant for a given reaction at a specific temperature, but will vary if the temperature changes.
• **Products vs Reactants**: A large K value (>1) suggests that products are favored at equilibrium, while a small K value (<1) indicates that reactants are favored.

In the context given, the equilibrium constant for methanol decomposition indicates the extent to which methanol (CH extsubscript{3}OH) decomposes into carbon monoxide (CO) and hydrogen gas ( extsubscript{2}H extsubscript{2}). After determining the equilibrium concentrations using the ICE table, these values are plugged into the K expression to find its numerical value at 250°C.

Methanol Decomposition

Methanol decomposition is a chemical reaction in which methanol (CH extsubscript{3}OH), a simple alcohol, breaks down into carbon monoxide (CO) and hydrogen gas ( extsubscript{2}H extsubscript{2}). This is a reversible reaction, as indicated by the double arrow, showing that the reaction can proceed in both directions until equilibrium is reached.

**Understanding Methanol Decomposition:**
- **Reaction Equation**: \[\mathrm{CH}_3\mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2\mathrm{H}_2(g)\]- **Stoichiometry**: For every one mole of methanol decomposed, one mole of carbon monoxide and two moles of hydrogen gas are produced.
- **Importance of Equilibrium**: At the equilibrium point, the rates of the forward and reverse reactions are equal. The concentration of each molecule remains constant over time, though not necessarily equal among different species.

Understanding this process is crucial for chemists and industries that rely on methanol as a feedstock for producing chemicals like formaldehyde, which itself serves as a precursor for many important industrial chemicals.