InterviewSolution
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InterviewSolution
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Calculate the equilibrium constant `K_(p)` and `K_(c )` for the reaction: `CO(g)+1//2O_(2)(g) hArr CO_(2)(g)`. Given that the partial pressure at equilibrium in a vessel at `3000 K` are `p_(CO)=0.4 atm, p_(CO_(2))=0.2 atm` |
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Answer» Correct Answer - B `CO(g)+1/2O_(2)(g)hArr CO_(2)(g)` `K_(p)=([p_(CO_(2))(g)])/([p_(CO)(g)][p_(O_(2))(g)]^(1//2))` `=0.6/(0.4xx(0.2)^(1//2))=0.6/(0.4xx0.4472)=3.356 "atm"^(-1//2)` `K_(p)=K_(c )(RT)^(Deltan)` `Deltan=1-(1+1/2)=-1/2` `K_(c )=K_(p)/((RT)^(Deltan))` `=3.356/((0.082xx3000)^(-1//2))` `=3.356xx(0.082xx3000)^(1//2)` `=3.356xx15.684=52.67` |
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