Ravnotezne koncentracije supstanci koje ucestvuju u sistemu
CO+H2O --- H2+CO2 su (mol/l): CO=0,02 ; H2O=0,32 ; H2=0,08; CO2=0,08.Kakve bi bile ravnotezne koncentracije ,ako bi se koncentracija CO povecala 4 puta?
CO+H2O<-->CO2+H2 Prvo izracunas K=(0.08*0.08)/(0.02*0.32)=1 Onda ide ovako: CO+H2O<-->CO2+H2 0.08-x..0.32-x...0.08+x..0.08+x 1=((0.08+x)(0.08+x))/((0.08-x)(0.32-x)) Resenje jednacine je x=0.0343M sto znaci da bi ravnotezne koncentracije bile: [CO]=0.0457M [H2O]=0.2857M [H2]=0.1143M [CO2]=0.1143M
"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it"Max Planck
CO+H2O<-->CO2+H2Prvo
CO+H2O<-->CO2+H2
Prvo izracunas K=(0.08*0.08)/(0.02*0.32)=1
Onda ide ovako:
CO+H2O<-->CO2+H2
0.08-x..0.32-x...0.08+x..0.08+x
1=((0.08+x)(0.08+x))/((0.08-x)(0.32-x))
Resenje jednacine je x=0.0343M sto znaci da bi ravnotezne koncentracije bile:
[CO]=0.0457M
[H2O]=0.2857M
[H2]=0.1143M
[CO2]=0.1143M
"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it" Max Planck