CH3OH can be prepared by CO + 2H2 f CH3OH Kc=4×10^2 At eqbm. 5L flask contains equal moles of CO and CH3OH . Hence the no. Of moles of H2 at eqbm is

Dear Student,

The computation are expressed below,


C = nV=n5 ltit is given moles of CO = CH3OH, if we take n =1 then the concentration of both CO and CH3OH are equal as below,[CO]=[CH3OH]=15=0.2 Mnow plugging the values in equilibrium equation we get,Kc=[CH3OH][CO][H2]24×102=[0.2][0.2][H2]2[H2]2=[0.2][0.2]×4×102=0.0025[H2]=0.05 MC=nVn=C×V=0.05 ×5 =0.25 mol

Regards.

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