Chapter 15, Problem 59GQ

### 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

# At 1800 K, oxygen dissociates very slightly into its atoms.O2(g) ⇄ 2 O(g)    KP= 1.2 × 10−10If you place 0.050 mol of O2 in a 10.-L vessel and heat it to 1800 K, how many O atoms are present in the flask?

Interpretation Introduction

Interpretation:

The number of oxygen atoms present in the flask in the dissociation of oxygen molecule into its atoms 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

Explanation

Given:

O2â€‰(g)â€‰â‡„â€‰2â€‰Oâ€‰(g)â€‰Kpâ€‰â€‰â€‰=â€‰1.2Ã—10-10Vâ€‰â€‰â€‰â€‰â€‰=â€‰10â€‰LTâ€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰1800â€‰KnO2â€‰â€‰â€‰â€‰â€‰â€‰=â€‰0.050

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

Kpâ€‰â€‰=â€‰Kcâ€‰(RT)Î”nâ€‰â€‰Kcâ€‰â€‰=â€‰â€‰Kp(RT)Î”nâ€‰â€‰=â€‰â€‰1.2Ã—10-10(0.082Ã—1800)2-1â€‰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€‰8.12Ã—10-13

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€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰O2â€‰(g)â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â‡„â€‰â€‰â€‰â€‰â€‰â€‰2â€‰Oâ€‰(g)Initialâ€‰concentrationâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰0

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