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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Lanthanum oxalate decomposes when heated to lanthanum(III) oxide, CO, and CO2.

La2(C2O4)3(s) ⇄ La2CO3(s) + 3 CO(g) + 3 CO2(g)

  1. (a) If, at equilibrium, the total pressure in a 10.0-L flask is 0.200 atm, what is the value of Kp?
  2. (b) Suppose 0.100 mol of La2(C2O4)3 was originally placed in the 10.0-L flask. What quantity of La2(C2O4)3 remains unreacted at equilibrium at 373 K?

a.

Interpretation Introduction

Interpretation:

The value of KP in the decomposition reaction of lanthanum oxalate and the quantity of lanthanum oxalate remain unreacted at equilibrium if initially 0.10mol of lanthanum oxalate is placed in the flask has to be given.

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

Explanation

To determine:

The value of KP in the decomposition reaction of lanthanum oxalate into lanthanum oxide, COandCO2

Given:

La2(C2O4)3(s)La2O3(s)+3CO(g)+3CO2(g)Totalpressure=0.2atmV=10L

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

KP=[CO]3[CO2]3

From the equation La2(C2O4)3(s)La2O3

(b)

Interpretation Introduction

Interpretation:

The value of KP in the decomposition reaction of lanthanum oxalate and the quantity of lanthanum oxalate remain unreacted at equilibrium if initially 0.10mol of lanthanum oxalate is placed in the flask has to be given.

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

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