The equilibrium mixture is found to contain 0.07,0.11,0.03 and 0.03 moles of CO,H2CH4 and H2O respectively for the reaction: CO(g) +3H2(g)O CH4(g)=H2O(g).
CO = 0,07
H2 = 0,11
CH4 = 0,03
H2O = 0,03
CO + 3H2 = CH4 + H2O
"K_{equ}= \\cfrac {[CH_4][H_2O]}{[CO][H_2]\u00b3}"
"K_{equ}= \\cfrac {0,03*0,03}{0,07*0,11\u00b3}=9,66"
Answer: Kequilibrium = 9,66
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