InterviewSolution
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InterviewSolution
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For the reaction `CO(g)+2H_(2)(g)hArrCH_(3)OH(g)` Hydrogen gas is introduced into a five-litre flask at `327^(@)C`, containing `0.2` mol of `CO(g)` and a catalyst, untill the pressure is `4.92 atm`. At this point, `0.1` mol of `CH_(3)OH(g)` is formed. Calculate the equilibrium constants `K_(p)` and `K_(c )`. |
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Answer» `{:(,,CO_((g)),+,2H_(2(g)),hArr,CH_(3)OH_((g))),(,"Initial mole",0.2,,a,,0),(,"Mole at equilibrium",(0.2-0.1),,(a-0.2),,0.1):}` Now total mole at equilibrium `=0.1+a-0.2+0.1=a` Also Mole `(n)=(PV)/(RT)=(4.92xx5)/(0.0821xx600)=0.499` `a=0.499` `:. [CH_(3)OH]=(0.1)/(5)`, `[CO]=(0.2-0.1)/(5)=(0.1)/(5)`, `[H_(2)]=(0.499-0.20)/(5)=(0.299)/(5)` `:. K_(c)=([CH_(3)OH])/([CO][H_(2)]^(2))=(0.1//5)/(0.1//5xx(0.299//5)^(2))` `=279.64 litre^(2)mol^(-2)` `K_(p)=K_(c)(RT)^(Deltan)` `=279.64xx(0.0821xx600)^(-2)` `=0.115 atm^(-2)` |
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