/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 110 A 4.72 -g sample of methanol (CH... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(250 . \mathrm{C} .\)

Short Answer

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

Step by step solution

01

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 \(\mathrm{CH}_3\mathrm{OH} = \frac{4.72 \text{g}}{32.042 \text{g/mol}} \approx 0.147\text{ moles}\).
02

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: \(\mathrm{CH}_3\mathrm{OH}(g)\): \((0.147 - x)\,\mathrm{mol/L}\) \(\mathrm{CO}(g)\): \(x\,\mathrm{mol/L}\) \(\mathrm{H}_2(g)\): \(2x\,\mathrm{mol/L}\)
03

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: \(\mathrm{CH}_3\mathrm{OH} = 0.147 - x \approx 0.141\,\mathrm{mol/L}\) \(\mathrm{CO} = x \approx \frac{2}{33}\,\mathrm{mol/L}\) \(\mathrm{H}_2 = 2x\approx \frac{4}{33}\,\mathrm{mol/L}\)
04

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 \(K \approx 7.94 \times 10^{-4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\) b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\)

At \(1100 \mathrm{K}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ Calculate the equilibrium partial pressures of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) produced from an initial mixture in which \(P_{\mathrm{SO}_{2}}=P_{\mathrm{O}_{2}}=\) 0.50 \(\mathrm{atm}\) and \(P_{\mathrm{so}_{3}}=0 .\) (Hint: If you don't have a graphing calculator, then use the method of successive approximations to solve, as discussed in Appendix \(1.4 . )\)

Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right),\) an extremely poisonous gas. Phosgene decomposes by the reaction $$\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C} .\) If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

In a given experiment, 5.2 moles of pure NOCl were placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

The gas arsine, \(\mathrm{AsH}_{3},\) decomposes as follows: $$2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g)$$ In an experiment at a certain temperature, pure \(\mathrm{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr. After 48 hours the pressure in the flask was observed to be constant at 488.0 torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(g)\) b. Calculate \(K_{\mathrm{p}}\) for this reaction.

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