Izračunajte vrijednost za konstantu ravnoteže reakcije A+3B=2C ako su ravnotežne koncentracije C(A)=C(C)=1,7*10 na -5 mol/L C(B)=0,025 mol/L
C(A)=c(B)=0,081 mol/L C(C)=0,0012 mol/L
A+3B----------------->2C
a) c(A)=c(C)=1.7*10(-5) mol/l c(B)=0.025 mol/l
K={C}rav.*2/({A}rav.*{B}rav.*3)
K=(1.7*10(-5)*2)/(1.7*10(-5)*(0.025)*3)=1.088
b) c(A)=c(B)=0.081 mol/l a c(C)=0.0012 mol/l
K=(0.0012)*2/(0.081*(0.081)*3)=27.88
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Hemijska ravnoteza
A+3B----------------->2C
a) c(A)=c(C)=1.7*10(-5) mol/l c(B)=0.025 mol/l
K={C}rav.*2/({A}rav.*{B}rav.*3)
K=(1.7*10(-5)*2)/(1.7*10(-5)*(0.025)*3)=1.088
b) c(A)=c(B)=0.081 mol/l a c(C)=0.0012 mol/l
K=(0.0012)*2/(0.081*(0.081)*3)=27.88