Standard Enthalpy of Hydration:标准的水化焓 热度: Enthalpy and entropy in chemical reactivity(化学反应的焓和熵) 热度: EnthalpyandEnthalpyChanges 1.1DefinitionofEnthalpy Review: Thefirstlawofthermodynamics:E=q+w 2 ΔV Zn Zn(s)+2H + (aq)Zn ...
1.1DefinitionofEnthalpy Review：➢Thefirstlawofthermodynamics：E=q+w ➢Whenopensystemwithonlymechanicalworkassociatedwithvolume：pressure-volumework,orP-Vwork ΔV w=Fs=-P*A*h=-PΔV Zn ΔV=Vfinal-Vntial ΔV>0,w<0 ΔV<0,w>0 H2(g)HCl(aq)Zn(s)+2H+(aq)Zn2+(aq)+H2(g)2 1.1...
FirstEA=-349kJNa+(g)+Cl-(g)L.E(Na++Cl-)=-788kJ Assumeenergy=0 Energy(kJ)Na-+Cl-(s)Buthowdoweget-788kJ?•Method1 •∆HatmNa+∆HatmCl∆+∆HIE∆+∆H1stAE+∆HL.E=∆Hformation•Rearrangeintermsof∆HL.E •Gives •∆HL.E=∆Hformation-(∆HatmNa+∆Hat...
Hformation Rearrange in terms of ?HL.E Gives ?HL.E= ?Hformation - (?HatmNa +?HatmCl? +?HIE? +?H1stAE) = -411 – (108 + 122 + 496+ (-349)) = -788kJmol-1 Method 2 – use the diagram Method 2 continued Add all the terms together – being careful with the signs. = ...
method 1hatmna +hatmcl +hie +h1stae+ hl.e= hformationrearrange in terms of hl.egives hl.e= hformation - (hatmna +hatmcl +hie +h1stae) = -411 (108 + 122 + 496+ (-349) = -788kjmol-1method 2 use the dia 4、gramenergy (kj)hat(na) = +108kjh(at) = 1/2cl2cl(g) =...
Rearrange in terms of ?HL.E ? Gives ? ?HL.E= ?Hformation - (?HatmNa +?HatmCl? +?HIE? +?H1stAE) ? = -411 – (108 + 122 + 496+ (-349)) ? = -788kJmol-1 Method 2 – use the diagram Energy (kJ) First I.E (Na) = +496kJ First EA = -349kJ ?H?(at) = 1/2Cl2...
FirstEA=-349kJNa+(g)+Cl-(g)L.E(Na++Cl-)=-788kJ Assumeenergy=0 Energy(kJ)Na-+Cl-(s)Buthowdoweget-788kJ?•Method1 •∆HatmNa+∆HatmCl∆+∆HIE∆+∆H1stAE+∆HL.E=∆Hformation•Rearrangeintermsof∆HL.E •Gives •∆HL.E=∆Hformation-(∆HatmNa+∆Hat...
Na+(g)+Cl-(g)Energy(kJ)L.E(Na++Cl-)=-788kJ Na-+Cl-(s)Buthowdoweget-788kJ?•Method1 •∆HatmNa+∆HatmCl∆+∆HIE∆+∆H1stAE+∆HL.E=∆Hformation•Rearrangeintermsof∆HL.E •Gives•∆HL.E=∆Hformation-(∆HatmNa+∆HatmCl∆+∆HIE∆+∆H1stAE)...
When activated carbon is placed in contact with a liquid there are different phenomena such as wetting, penetration of the molecules into the pores, and the formation of a layer that involves specific interactions; these can be studied by immersion calorimetry, because the heat exchanged during ...