   Chapter 15, Problem 62GQ

Chapter
Section
Textbook Problem

The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 800 °C.CaCO3(s) ⇄ CaO(s) + CO2(g) (a) What is Kc for the reaction? (b) If you place 22.5 g of CaCO3 in a 9.56-L container at 800 °C, what is the pressure of CO2 in the container? (c) What percentage of the original 22.5-g sample of CaCO3 remains undecomposed at equilibrium?

(a)

Interpretation Introduction

Interpretation:

The value of Kc in the decomposition reaction of calcium carbonate has to be determined.

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.

Kp=Kc(RT)ΔnKc-EquilibriumconstantintermsofconcentrationKp-EquilibriumconstantintermsofpressureΔn-changeinnumberofmoles

Molarity=NumberofmolesVolumeofsolution(L)

Explanation

Given:

CaCO3(s)CaO(s)+CO2(g)KP=1.16T=800°C =1073K

Using the equation Kp=Kc(RT)Δn we can find out the value of Kc

Kp

(b)

Interpretation Introduction

Interpretation:

The pressure of carbon dioxide in the container 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.

Kp=Kc(RT)ΔnKc-EquilibriumconstantintermsofconcentrationKp-EquilibriumconstantintermsofpressureΔn-changeinnumberofmoles

Molarity=NumberofmolesVolumeofsolution(L)

Interpretation Introduction

Interpretation:

The percentage of original 22.5 g sample of CaCO3 remain undecomposed at equilibrium 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.

Kp=Kc(RT)ΔnKc-EquilibriumconstantintermsofconcentrationKp-EquilibriumconstantintermsofpressureΔn-changeinnumberofmoles

Molarity=NumberofmolesVolumeofsolution(L)

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