Chapter 15, Problem 53GQ

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

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

# 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.)

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

Explanation

Given:

2NO2â€‰(g)â€‰â‡„â€‰â€‰N2O4â€‰(g)KPâ€‰â€‰â€‰â€‰=â€‰7.1Vâ€‰â€‰â€‰â€‰=â€‰â€‰15â€‰LTâ€‰â€‰â€‰=â€‰300â€‰K

Using the equation Kpâ€‰â€‰=â€‰Kcâ€‰(RT)Î”nâ€‰â€‰, we can find the value of Kc

Kpâ€‰â€‰=â€‰Kcâ€‰(RT)Î”nâ€‰â€‰Kcâ€‰â€‰=â€‰â€‰Kp(RT)Î”nâ€‰â€‰=â€‰â€‰7.1(0.082Ã—300)-1â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰174.87

For the reaction 2NO2â€‰(g)â€‰â‡„â€‰â€‰N2O4â€‰(g)

Kcâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰[N2O4][NO2]2174.87â€‰=â€‰â€‰xy2â€‰xâ€‰â€‰â€‰â€‰â€‰=â€‰174.87â€‰y2â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰Equationâ€‰â€‰â€‰â€‰(1)â€‰

Molarityâ€‰ofâ€‰N2O4â€‰â€‰=â€‰No.â€‰ofâ€‰molesVolumeâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰xâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰nN2O415â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰nN2O4â€‰â€‰â€‰â€‰â€‰=â€‰15xâ€‰â€‰â€‰

â€‰nN2O4â€‰â€‰=â€‰â€‰mN2O4Molarâ€‰massâ€‰ofâ€‰â€‰N2O4mN2O4â€‰â€‰â€‰=â€‰â€‰nN2O4â€‰Ã—â€‰Molarâ€‰massâ€‰ofâ€‰â€‰N2O4â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰15xÃ—92â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰1380x

mNO2â€‰â€‰â€‰=â€‰â€‰nNO2â€‰Ã—â€‰Molarâ€‰massâ€‰ofâ€‰â€‰NO2â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰15yÃ—46â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰690y

Total mass is given as 6

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