review for equilibrium test：平衡试验综述.doc

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1、Answers：EQUILIBRIUMPRACTICETEST1.Thedissociationofammoniaat27°ChasaKeqvalueof2.63x10'9.Heat+2NH3(g)0N2(g)+3H2(g)concentrationsofa)If1.00Mammoniaisplacedinareactionvessel,calculatetheequilibrN2andH2.2NH3(g)<=>N2(g)+3出g)Keq二2.63xIO9[I]1.0000」C1・2x+3xx3x[E]1.00

2、-2x-2xisinsignificantbecause1.00>lOOOxKeqK’q二IN2IH2f[NH3]22.63xIO9=(x)(33(1.2.63xxX二0.00314The2二0.00314Mand[H訂二3(0.00314)二0.00942MpercentyieldofN2(g)?Ifthereactioncouldgotocompletion,1.00molofNH3couldproduce0.500molN2.(2:1reaction)%yield二0.00314x100%二0.628%0

3、.500Increasethetemperature(reactionisendo+hermic)RemoveeitherN2orH2Decreasethepressurebyincreasingthevolumeofthecontainereleft.1.0.20moleseachofA#BandCaremixedintoa2.0Lcontainerandallowedtoreactaccordingtothefollowingequation：A(g)+C(g)Keq二l・0x106.Calculateth

4、eequilibriumconcentrationsofeachgas・A(g)+B(g)《TC(g)Keq=1.0X106[I]0.100.100.10「C]+x+x-x(swenotebelow)[E]0.10+x0.10+x0.10-xQ二[C]二0.10二10[A][B](0.10)(0.10)Q>Keqthereforethereactionneeds喝二©二0.10-x[A][B](0.10+x)(0.10>x)l.OxlO6(0.010+0.20小l.OxlCT*+ZOxlOMl.Oxlikete

5、rmsarecollectedl.OxlOV+x-0.10=0^X=-l±/(I4(1・0x10・6)(Q10)X=OjTherefore[A]^0.10+0.10二0.20M[B]10.20M[C]=0.10-0.10二0(Thisisntpossible.Weknowthatthevalueofx<0.10but汁hasbeenroundedto0.10onthecalculator.Togetthenon-zerovalueofC,weneedtoresolveforCusingourequilibriu

6、mvaluesofAandB.l.OxlO6=[C](0.20)(0.20)Therefore[C]=4.0xl0_8M1.Considerthefollowingequilibrium：NH3(aq)+H20(l)《TNH/(aq)+OHaq)1.00NH3(aq)[I]Stress0[E]+x1.00+xAtequilibrium,theconcentrationsofNH3(Qq),NH4&)andOH(aq)are1.00M,0.0010Mand0.018Mrespectively.To500.0mL

7、ofthisequilibriummixture,0.050molofOHareadded.CalculatethenewequilibriumconcentrationsofNH4+andOH；0.0010-x0FU8-XKeq二INH匸g口二(0.0010)(0.018)二1.8>aWf(frominitialequilibrium)[NH3]―'1.8x10乞(0.0010・x)(0.118-x)二n8xio-4-0.119x+x*2*1.00+x(insignificantsincex<0.0010)

8、X2-0.119x+l.OxlO4=0X二0・119±/(0.11V・4⑴(1OxlOQ2X二8.46x10"Therefore[NH4+]010-0.000846=1.54xlO'4M]二0.118・0.000846二0.U7M