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`64 gm` of `CH_(4) "and" 68gm "of" H_(2)S` was placed in an close container and heated up to `727^(@)C` following equilibrium is established in gaseous phase reaction is: `CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` The total pressure at equilibrium is `1.6` atm and partial pressure of `H_(2) "is" 0.8` atm. ThenA. Total moles at equilibrium `4.8`B. `K_(P)=K_(C)(RT)^(2)`C. Mole fraction `H_(2)` at equilibrium`=0.5`D. On increasing moles of `H_(2)S` equilibrium constant increases. |
Answer» Correct Answer - A::B::C::D `CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` `{:(t=0,64gm,68gm),("Moles",4"mole",2"mole"):}` `4-x 2-2x x 4x` Total moles at equilibrium`=4-x+2-2x+x+4x=(6-2x)` Acc. To `PV=nRT,T=727+273=1000K` `1.84xxV=(6-2x)xxRxx1000 `....(1) For `H_(2) "gas" V,P_(H_(2))=n_(H_(2))RTto0.8xxV=4xxRxx1000` ....(2) From eq (1) and(2) `(1.6)/(0.8)=((6-2x))/(4x)` `x=0.6` |
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