Hemijske veze pomoc

Submitted by andjela on Tue, 10/29/2013 - 20:57
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P(g) +3Cl(g)-> PCl3(g) ,delta H= -983 kJ/mol
P(s) +1,5 Cl2(g) ->PCgl3(), delta H= -305 kJ/mol
P(g) + 3H(g)-> PH3(g) , delta H = -958 kJ/mol
P(s)+1,5H2()g ->PH3(g),deltA H = -8,4
P(s)-> P(g),delta H= 314 kJ/mol
Koristeći gore navedene informacije izračunati prosječne vezivne energije sledećih veza ?
a)P-Cl u PCl3
b)Cl-Cl u Cl2
c)P-H u PH3
d)H-H u H2

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milicaradenkovic - Wed, 10/30/2013 - 08:50

Jel imas resenje zadatka?

andjela's picture

andjela - Wed, 10/30/2013 - 20:37

nemam :/

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milicaradenkovic - Wed, 10/30/2013 - 22:33

Ja mislim mada nisam sigurna da se zadatak radi na sledeci nacin

a) 2PCl3(g)------------------->6Cl(g)+2P(s)

deltaH= 983+305-314= 974 kJ/mol

deltaH=6*(P-Cl)

P-Cl= 974/6= 162.3

b) PCl3(g)------------>P(g)+3Cl(g)

P(s)+1.5Cl(g)------------->PCl3(g)

P(g)--------------->P(s)kada se jednacine saberu dobija se 

1.5Cl2(g)---------------->3Cl(g)

deltaH=983-305-314=364 kJ/mol

Cl-Cl=364/3=121.33

c) 2PH3(g)---------------->2P(g)+6H(g)

deltaH=(958+8.4+314)=1280.4 kJ/mol

deltaH=6*(P-H)

1280.4=6*(P-H)

P-H=1280.4/6=213.4

d)PH3(g)--------------.P(g)+3H(g)

P(s)+1.5H2(g)---------------.PH3(g)

P(g)------------->P(s) kada se saberu jednacine dobija se 

1.5H2(g)-------------->3H(g)

deltaH=958-8.4-314=635.6

H-H=635.6/3=211.87 b i d mislim da je tacno a ti proveri a i c