Toplota

Submitted by lupica on Sat, 04/04/2015 - 13:12
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Pri sagorevanju 6.538g cinka u cistom kiseoniku oslobadja se 34.82 kJ toplote. Izracunaj standardnu reakcionu toplotu tog procesa

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milicaradenkovic - Sat, 04/04/2015 - 14:01

2Zn+O2----------------->2ZnO

n(Zn)=m/Mr=6.538/65=0.10058 mol

n(Zn)1=n(Zn)/2=0.10058/2=0.05029 mol

deltaH=Q/n=34.82/0.05029=692.384 kJ/mol

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lupica - Sat, 04/04/2015 - 14:54

resenje je u zbirci -348.2

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milicaradenkovic - Sat, 04/04/2015 - 15:32

Obicno se uzima po jednom molu entalpija ali su oni uzeli po 2 mola pa je zato resenje -348.2 kJ/mol

deltah=Q/n(Zn)= -34.82/0.1= -348.2 kJ/mol