   Chapter 15, Problem 75SCQ

Chapter
Section
Textbook Problem

Suppose a tank initially contains H2S at a pressure of 10.00 atm and a temperature of 800 K. When the reaction has come to equilibrium, the partial pressure of S2 vapor is 0.020 atm. Calculate Kp.2 H2S(g) ⇄ 2 H2(g) + S2(g)

Interpretation Introduction

Interpretation:

The KP for the reaction when dissociation of H2S at a pressure of 10atm 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

Explanation

Given:

2H2S(g)2H2+S2(g)PS2=0.020P=10atmT=800K

For the reaction 2H2S(g)2H2+S2(g)

2H2S(g)2H2+S2(g)Initialpressure1000Change-2x2xxEquilibriumpressure10-2x2xx

PS2=x=0

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