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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

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:

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

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