/*! 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 58 A limestone decomposes at high t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A limestone decomposes at high temperatures. $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ At \(1000^{\circ} \mathrm{C}, K_{\mathrm{p}}=3.87 .\) If pure \(\mathrm{CaCO}_{3}\) is placed in a 5.00 -L flask and heated to \(1000^{\circ} \mathrm{C},\) what quantity of \(\mathrm{CaCO}_{3}\) must decompose to achieve the equilibrium pressure of \(\mathrm{CO}_{2} ?\)

Short Answer

Expert verified
0.1854 moles of \( \mathrm{CaCO}_3 \) must decompose.

Step by step solution

01

Write the equilibrium expression

Since the decomposition involves only gaseous products, the equilibrium expression is related to the partial pressure of COâ‚‚. Given \( K_p = 3.87 \) at \( 1000^{\circ} C \), and the reaction: \[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]The equilibrium constant expression is:\[ K_p = P_{\mathrm{CO}_2} \] Therefore, \( P_{\mathrm{CO}_2} = 3.87 \) atm.
02

Relate COâ‚‚ pressure to moles

Using the Ideal Gas Law \( PV = nRT \), we calculate the number of moles \( n \) of \( \mathrm{CO}_2 \) at equilibrium:\[ n = \frac{P \cdot V}{R \cdot T} \]Here, \( P = 3.87 \) atm, \( V = 5.00 \) L, \( R = 0.0821 \ \frac{L \, atm}{mol \, K} \), and \( T = 1000^{\circ}C + 273.15 = 1273 \) K.
03

Substitute values into the Ideal Gas Law

Calculate the moles of \( \mathrm{CO}_2 \): \[ n = \frac{3.87 \times 5.00}{0.0821 \times 1273} \]Calculate the exact value to find the moles of \( \mathrm{CO}_2 \) formed.
04

Solve for moles of COâ‚‚

Upon calculating, we find:\[ n = \frac{19.35}{104.4133} \approx 0.1854 \text{ moles of } \mathrm{CO}_2 \]This is the amount of \( \mathrm{CO}_2 \) produced at equilibrium, which corresponds to the moles of \( \mathrm{CaCO}_3 \) decomposed.

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 91Ó°ÊÓ!

Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of gases under various conditions. It relates pressure, volume, and temperature of a gas to the number of moles present. The equation is expressed as:\[ PV = nRT \]where:
  • P stands for pressure in atmospheres (atm)
  • V is the volume in liters (L)
  • n is the number of moles of the gas
  • R is the ideal gas constant, approximately 0.0821 L·atm/mol·K
  • T is the temperature in Kelvin (K)
To use this law effectively, always convert the temperature to Kelvin by adding 273.15 to the Celsius temperature. The Ideal Gas Law helps predict how a gas will respond to changes in pressure, volume, or temperature when other variables are held constant. In the context of equilibrium reactions involving gases like the decomposition of limestone, the Ideal Gas Law can be used to calculate how much COâ‚‚ is produced in a reaction vessel by linking pressures to moles through the equation.
Equilibrium Constant
In chemical reactions, the equilibrium constant (\( K_p \) or \( K_c \)) quantifies the ratio of product concentrations to reactant concentrations at equilibrium for a reaction. For reactions involving gases, the expression can be formulated in terms of partial pressures as \( K_p \). For the decomposition of limestone:\[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]The equilibrium constant \( K_p \) is defined only for the gaseous components:\[ K_p = P_{\mathrm{CO}_2} \]This equation indicates that at equilibrium, the partial pressure of COâ‚‚ is equal to \( 3.87 \) atm as provided in the problem statement.

This value of \( K_p \) tells us how far the reaction proceeds towards products under specified conditions. A larger \( K_p \) suggests a significant formation of products, while a smaller \( K_p \) indicates a reaction that hardly favors products.
Decomposition Reaction
Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. The general form can be expressed as follows:\[ \text{AB} \rightarrow \text{A} + \text{B} \]Decomposition reactions require energy, often in the form of heat, to break bonds. In the limestone decomposition:\[ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]The limestone (\( \mathrm{CaCO}_{3} \)) decomposes into calcium oxide (\( \mathrm{CaO} \)) and carbon dioxide gas (\( \mathrm{CO}_{2} \)).

This type of reaction is important in industrial processes, such as cement production. At high temperatures, the forward reaction occurs, resulting in the release of COâ‚‚ gas. Understanding the conditions for equilibrium in such reactions helps optimize processes to either enhance or minimize the formation of specific products.

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

The equilibrium constant, \(K_{c}\) for the following reaction is 1.05 at \(350 \mathrm{K}\). $$ 2 \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CCl}_{4}(\mathrm{g}) $$ If an equilibrium mixture of the three gases at \(350 \mathrm{K}\) contains \(0.0206 \mathrm{M} \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) and \(0.0163 \mathrm{M}\) \(\mathrm{CH}_{4},\) what is the equilibrium concentration of \(\mathrm{CCl}_{4} ?\)

Write equilibrium constant expressions for the following reactions. For gases, use either pressures or concentrations. (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\) (b) \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})\) (c) \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\) (d) \(\mathrm{NiO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Ni}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\)

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

The size of a flask containing colorless \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) and brown \(\mathrm{NO}_{2}(\mathrm{g})\) at equilibrium is rapidly reduced to half the original volume. $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \approx 2 \mathrm{NO}_{2}(\mathrm{g})$$ (a) What color change (if any) is observed immediately upon halving the flask size? (b) What color change (if any) is observed during the process in which equilibrium is reestablished in the flask?

Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide. $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) $$ (a) Laboratory measurements at \(986^{\circ} \mathrm{C}\) show that there are 0.11 mol each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) vapor and 0.087 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) at equilibrium in a 50.0 -L container. Calculate the equilibrium constant for the reaction at \(986^{\circ} \mathrm{C}\) (b) Suppose 0.010 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) are placed in a 200.0 -L. container. When equilibrium is achieved at \(986^{\circ} \mathrm{C},\) what amounts of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) in moles, would be present? [Use the value of \(K_{c}\) from part (a).]

See all solutions

Recommended explanations on Chemistry Textbooks

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.