Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Textbook Question
Chapter 15, Problem 53GQ

A 15-L flask at 300 K contains 6.44 g of a mixture of NO2 and N2O4 in equilibrium. What is the total pressure in the flask? (Kp for 2 NO2 (g) ⇄ N2O4(g) is 7.1.)

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The total pressure in the flask that contain 6.44g mixture of NO2andN2O4 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

Ideal gas equation:

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

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

Answer to Problem 53GQ

The total pressure in the flask is 0.13atm.

Explanation of Solution

Given:

2NO2(g)N2O4(g)KP=7.1V=15LT=300K

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

Kp=Kc(RT)ΔnKc=Kp(RT)Δn=7.1(0.082×300)-1=174.87

For the reaction 2NO2(g)N2O4(g)

Kc=[N2O4][NO2]2174.87=xy2x=174.87y2Equation(1)

MolarityofN2O4=No.ofmolesVolumex=nN2O415nN2O4=15x

nN2O4=mN2O4MolarmassofN2O4mN2O4=nN2O4×MolarmassofN2O4=15x×92=1380x

mNO2=nNO2×MolarmassofNO2=15y×46=690y

Total mass is given as 6.44g

1380x+690y=6.44g690(2x+y)=6.44(2x+y)=0.0093y=0.0093-2xEquation(2)

Substituting (2) into (1),

x=174.87×y2=174.87×(0.0093-2x)2Solvingthiswegetx=0.0038

y=0.0093-2×0.0038=0.0017

nN2O4=15x=15×0.0038=0.057

nNO2=15y=15×0.0017=0.025

Totalnumberofmoles=0.057+0.025=0.082

According to ideal gas equation

PV=nRTP=nRTV=0.082×0.082×30015=0.13atm

Conclusion

The total pressure in the flask that contain 6.44g mixture of NO2andN2O4 was calculated

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Chapter 15 Solutions

Chemistry & Chemical Reactivity

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