   Chapter 15, Problem 58GQ

Chapter
Section
Textbook Problem

Limestone decomposes at high temperatures.CaCO3(s) ⇄ CaO(s) + CO2(g)At 1000 °C, Kp = 3.87. If pure CaCO3 is placed in a 5.00-L flask and heated to 1000 °C, what quantity of CaCO3 must decompose to achieve the equilibrium pressure of CO2?

Interpretation Introduction

Interpretation:

The quantity of calcium carbonate to achieve the equilibrium pressure of CO2 in the decomposition reaction has to be calculated.

Concept Introduction:

Equilibrium constant in terms of pressure[Kp]: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

aA(g)+bB(g)cC(g)+dD(g)KP=PCc×PDdPAa×PBb

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

Ideal gas equation:

PV=nRTP-PressureV-VolumeR-UniversalgasconstantT-Temperaturen-numberofmoles

Explanation

Given:

CaCO3(s)CaO(s)+CO2Kp=3.87V=5LT=1000°C

In the given reaction calcium carbonate and calcium oxide are in solid state and thus its pressure can be taken as unity. There is no role of calcium carbonate and calcium oxide in the equilibrium expression.

For this equation:

CaCO3(s)CaO(s)+CO2

Kp=[PCO2][SinceP(CaCO3)andP(CaO)hasnorole]3

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