Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 82

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

Short Answer

Expert verified
For the given mixtures: (a) The amount of \(CaCO_3\) will increase. (b) The amount of \(CaCO_3\) will decrease. (c) The amount of \(CaCO_3\) will remain the same.

Step by step solution

01

Calculate the moles of given reactants and products for each case

Use the molar mass of CaCO3, CaO, and CO2 to determine the moles of each in every case. Then, divide the moles by the volume to obtain their respective concentrations. Remember that solids do not affect the equilibrium situation, so solid products and reactants can be ignored when calculating Q. (a) Moles and Concentrations: \(n_{CaCO3} = \frac{15 g}{100.09 \frac{g}{mol}} = 0.15 mol\) \(C_{CaCO3} = \frac{0.15 mol}{10.0 L} = 0.015 M\) \(n_{CaO} = \frac{15 g}{56.08 \frac{g}{mol}} = 0.267 mol\) \(C_{CaO} = \frac{0.267 mol}{10.0 L} = 0.0267 M\) \(n_{CO2} = \frac{4.25 g}{44.01 \frac{g}{mol}} = 0.0965 mol\) \(C_{CO2} = \frac{0.0965 mol}{10.0 L} = 0.00965 M\) Repeat the same process for cases (b) and (c).
02

Calculate the reaction quotient (Q) for each case

According to the balanced chemical equation: \(Q_c = \frac{[CO_2]}{[CaCO_3][CaO]} \) To calculate Q for each case, substitute the values calculated in Step 1: (a) \(Q_{c(a)} = \frac{0.00965}{(0.015)(0.0267)} = 0.0239\) Repeat the same calculation for cases (b) and (c).
03

Compare Q and Kc for each case to determine the direction of the reaction

In each case, compare Q to the given equilibrium constant, Kc. If \(Q < K_c\), the system will shift forward to reach equilibrium and thus, the amount of CaCO_3 will decrease. If \(Q > K_c\), the system will shift backward to reach equilibrium and the amount of CaCO_3 will increase. If \(Q = K_c\), the system is at equilibrium, and the amount of CaCO_3 will remain the same. (a) Since \(Q_{c(a)} = 0.0239 > K_c = 0.0108\), the reaction will shift backward, and the amount of \(CaCO_3\) will increase. Now, compare Q and Kc for cases (b) and (c).
04

Give the conclusion for each case

Summarize the results for each case: (a) The amount of \(CaCO_3\) will increase. (b) Calculate and compare Q and Kc to determine whether the amount of \(CaCO_3\) will increase, decrease, or remain the same. (c) Calculate and compare Q and Kc to determine if the amount of \(CaCO_3\) will increase, decrease, or remain the same.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$\begin{aligned} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) & \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c} &=2.0 \\ 2 \mathrm{NO}(g) & \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c} &=2.1 \times 10^{30} \end{aligned}$$

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\) (f) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (g) \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}^{-}(a q)\)

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

A mixture of \(0.10 \mathrm{~mol}\) of \(\mathrm{NO}, 0.050 \mathrm{~mol}\) of \(\mathrm{H}_{2},\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(1.0-\mathrm{L}\) vessel at \(300 \mathrm{~K}\). The following equilibrium is established: $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ At equilibrium \([\mathrm{NO}]=0.062 \mathrm{M}\). (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\). (b) Calculate \(K_{c}\)

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

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.

Sign-up for free