Question

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).

Answer 1

CO = 0,07

H₂ = 0,11

CH₄ = 0,03

H₂O = 0,03

CO + 3H₂ = CH₄ + H₂O

$K_{equ}= \cfrac {[CH_4][H_2O]}{[CO][H_2]³}$

$K_{equ}= \cfrac {0,03*0,03}{0,07*0,11³}=9,66$

Answer: K_equilibrium = 9,66