CH3OH can be prepared by CO + 2H2 f CH3OH Kc=4×10^2 At eqbm 5L flask contains equal moles of - Chemistry - Chemical and Ionic Equilibrium

Question

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

Yasodharan · Accepted Answer

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=nV⇒n=C×V=0 05 ×5 =0
25 mol

Regards

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