/*! 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 12 You place 3.00 mol of pure \(\ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 12

You place 3.00 mol of pure \(\mathrm{SO}_{3}\) in an \(8.00-\mathrm{L}\). flask at \(1150 \mathrm{K}\). At equilibrium, 0.58 mol of \(\mathrm{O}_{2}\) has been formed. Calculate \(K_{c}\) for the reaction at \(1150 \mathrm{K}\). $$ 2 \mathrm{sO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

Short Answer

Expert verified
\(K_c \approx 0.0288\)

Step by step solution

01

Understand the Reaction

The balanced chemical reaction is \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\). This equilibrium involves the decomposition of sulfur trioxide (\(\mathrm{SO}_3\)) into sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen gas (\(\mathrm{O}_2\)).
02

Determine Initial Amounts and Changes

Initially, we have 3.00 mol of \(\mathrm{SO}_3\). No \(\mathrm{SO}_2\) or \(\mathrm{O}_2\) is present initially. If 0.58 mol of \(\mathrm{O}_2\) is formed at equilibrium, then by stoichiometry, 1.16 mol (\(2 \times 0.58\) mol) of \(\mathrm{SO}_3\) is consumed to produce 1.16 mol of \(\mathrm{SO}_2\).
03

Calculate Moles at Equilibrium

At equilibrium, the moles of each substance are:- \(\mathrm{SO}_3\): \(3.00 - 1.16 = 1.84\) mol- \(\mathrm{SO}_2\): \(1.16\) mol- \(\mathrm{O}_2\): \(0.58\) mol
04

Calculate Concentrations

Since the volume of the flask is 8.00 L, we calculate the concentrations as follows:- \([\mathrm{SO}_3]\): \(\frac{1.84}{8.00} = 0.230\,\mathrm{mol/L}\)- \([\mathrm{SO}_2]\): \(\frac{1.16}{8.00} = 0.145\,\mathrm{mol/L}\)- \([\mathrm{O}_2]\): \(\frac{0.58}{8.00} = 0.0725\,\mathrm{mol/L}\)
05

Write the Expression for the Equilibrium Constant, \(K_c\)

The equilibrium constant expression for the reaction is:\[ K_c = \frac{([\mathrm{SO}_2]^2 [\mathrm{O}_2])}{[\mathrm{SO}_3]^2} \]
06

Calculate \(K_c\)

Substitute the equilibrium concentrations into the expression:\[ K_c = \frac{(0.145)^2 \times 0.0725}{(0.230)^2} \]Calculate:1. \((0.145)^2 = 0.021025\)2. \(0.021025 \times 0.0725 = 0.0015253125\)3. \((0.230)^2 = 0.0529\)4. Finally, \(K_c = \frac{0.0015253125}{0.0529} \approx 0.0288\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the concentrations of the reactants and products stop changing with time. This happens because the rates of the forward and reverse reactions become equal. In the context of the given exercise, the reaction involves the decomposition of sulfur trioxide (\(\mathrm{SO}_3\)) into sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen (\(\mathrm{O}_2\)).
When the system reaches equilibrium, the quantity of each substance remains constant, though not necessarily equal between reactants and products. Here, 0.58 mol of \(\mathrm{O}_2\) is formed, signaling that equilibrium has been achieved.
It is crucial to understand that equilibrium does not mean the reactants and products are in equal concentrations. It simply means their concentrations are stable over time.
Outcomes of chemical equilibrium are described by equilibrium constants like \(K_c\), which help us understand the relative quantities of products and reactants at equilibrium.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It uses the coefficients from the balanced chemical equation to determine how much of each substance is consumed or produced. In this exercise, the reaction is given as: \[2 \mathrm{SO}_{3}(\text{g}) \rightleftharpoons 2 \mathrm{SO}_2(\text{g}) + \mathrm{O}_2(\text{g})\]. This balanced equation shows us that two moles of \(\mathrm{SO}_3\) decompose to form two moles of \(\mathrm{SO}_2\) and one mole of \(\mathrm{O}_2\).
  • For every mole of \(\mathrm{O}_2\) produced, 2 moles of \(\mathrm{SO}_3\) are consumed.
  • Similarly, each mole of \(\mathrm{SO}_2\) formed corresponds to one mole of \(\mathrm{SO}_3\) degraded.
By understanding stoichiometry, we calculated that 0.58 mol of \(\mathrm{O}_2\) production corresponds to 1.16 mol of \(\mathrm{SO}_3\) being consumed and an equal amount of \(\mathrm{SO}_2\) being produced. This step is fundamental to proceeding with the calculation of equilibrium concentrations.
Equilibrium Concentration
Equilibrium concentration refers to the amount of each reactant and product present in a system at equilibrium. To find these concentrations, it is crucial to know the initial amounts and any changes that occur as the system reaches equilibrium.
In this exercise, initially, we have 3.00 moles of \(\mathrm{SO}_3\). No \(\mathrm{SO}_2\) or \(\mathrm{O}_2\) is present at the start. Through the reaction stoichiometry, we determine that when equilibrium is reached, the moles of each substance are: 1.84 mol of \(\mathrm{SO}_3\), 1.16 mol of \(\mathrm{SO}_2\), and 0.58 mol of \(\mathrm{O}_2\).
  • The concentration \([\mathrm{SO}_3]\) at equilibrium is \(\frac{1.84}{8.00} = 0.230\,\mathrm{mol/L}\)
  • The concentration \([\mathrm{SO}_2]\) at equilibrium is \(\frac{1.16}{8.00} = 0.145\,\mathrm{mol/L}\)
  • The concentration \([\mathrm{O}_2]\) at equilibrium is \(\frac{0.58}{8.00} = 0.0725\,\mathrm{mol/L}\)
These concentrations are then used to calculate the equilibrium constant \(K_c\), providing insight into the reaction's dynamics at that specific temperature and pressure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Lanthanum oxalate decomposes when heated to lanthanum (III) oxide, \(\mathrm{CO},\) and \(\mathrm{CO}_{2}\) \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}(\mathrm{s}) \rightleftharpoons \mathrm{La}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g})+3 \mathrm{CO}_{2}(\mathrm{g})\) (a) If, at equilibrium, the total pressure in a \(10.0-\mathrm{L}\). flask is 0.200 atm, what is the value of \(K_{p} ?\) (b) Suppose 0.100 mol of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\) s was originally placed in the \(10.0-\mathrm{L}\). flask. What quantity of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\) remains unreacted at equilibrium at \(373 \mathrm{K}^{2}\)

Hemoglobin (Hb) can form a complex with both \(\mathrm{O}_{2}\) and CO. For the reaction $$ \mathrm{HbO}_{2}(\mathrm{aq})+\mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{HbCO}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g}) $$ at body temperature, \(K\) is about \(200 .\) If the ratio \([\mathrm{HbCO}] /\left[\mathrm{HbO}_{2}\right]\) comes close to \(1,\) death is probable. What partial pressure of \(\mathrm{CO}\) in the air is likely to be fatal? Assume the partial pressure of \(\mathrm{O}_{2}\) is \(0.20 \mathrm{atm}\).

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), is used as a reagent in the synthesis of organic compounds. When heated to a sufficiently high temperature, it decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\). \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{c}=0045 \mathrm{at} 375^{\circ} \mathrm{C}\) (a) A \(1.00-1 .\) flask containing \(6.70 \mathrm{g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is heated to \(375^{\circ} \mathrm{C}\). What is the concentration of each of the compounds in the system when equilibrium is achieved? What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has dissociated? (b) What are the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2},\) and \(\mathrm{Cl}_{2}\) at equilibrium in the \(1.00-\mathrm{L}\). flask at \(375^{\circ} \mathrm{C}\) if you begin with a mixture of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(6.70 \mathrm{g})\) and \(\mathrm{Cl}_{2}\) \((1.00 \mathrm{atm}) ?\) What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has dissociated? (c) Compare the fractions of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in parts (a) and (b). Do they agree with your expectations based on Le Chatelier's principle?

Calculate \(K\) for the reaction $$ \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) $$ given the following information: $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) & K=1.6 \\ \mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) & K=0.67 \end{array} $$

Consider the following equilibria involving \(\mathrm{SO}_{2}(\mathrm{g})\) and their corresponding equilibrium constants. \(\mathrm{SO}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{SO}_{3}(\mathrm{g}) \quad K_{1}\) \(2 \mathrm{sO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{2}\) Which of the following expressions relates \(K_{1}\) to \(K_{2} ?\) (a) \(K_{2}=K_{1}^{2}\) (d) \(K_{2}=1 / K_{1}\) (b) \(K_{2}^{2}=K_{1}\) (e) \(K_{2}=1 / K_{1}^{2}\) (c) \(K_{2}=K_{1}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.