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For the reaction C_(2)H_(6)(g)hArrC_(2)H_(4)(g)+H_(2)(g),DeltaG^(@)=22.38KJ mol^(-1)" at "900K. If pure C_(2)H_(6) is passed over a suitable catalyst at a temperature of 900 K and a pressure of 1.0 atmosphere, then calculate X. Where X=100xx"mole percent of hydrogen present at equilibrium". [Antilog(1.299=0.05)] |
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Answer» Or `K_(p)="antilog"[-(DeltaG^(@))/(2.303RT)]="antilog"[-(22.38xx10^(3)J)/((2.303)(8.314)(900))]` `K_(p)="antilog"[-1.299]` `K_(p)=0.050` `{:("Now,"""C_(2)H_(6)(g)hArrC_(2)H_(4)(g)+H_(2)(g)),("At equilibrium,"""(C-Calpha)""Calpha""Calpha):}` where 'C' is the initial concentration of `C_(2)H_(6).` Total moles at equilibrium `=C-Calpha+Calpha+Calpha=C+Calpha=C(1+alpha)` THEREFORE, the partial PRESSURE of `C_(2)H_(6)=(1-alpha)/(1+alpha).P` partial pressure of `H_(2)=(alpha)/(1+alpha).P` `K_(P)=0.050=([(alpha)/(1+alpha).P]^(2))/([(1-alpha)/(1+alpha)].P)=(alpha^(2))/((1+alpha)(1-alpha)).P=(alpha^(2))/(1-alpha^(2)).P` Then, `K_(p)=0.050=(alpha^(2))/(1+alpha^(2)).1" or "alpha=0.2182` Mole percent of `H_(2)=(alpha)/(1+alpha)xx100=(0.2182)/(1.2182)xx100=17.91` `X=17.91xx100=1791` Ans. |
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