InterviewSolution
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InterviewSolution
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The equilibrium constant `(K_(p))` for the decomposition of gaseous `H_(2)O` `H_(2)O(g)hArr H_(2)(g)+(1)/(2)O_(2)(g)` is related to the degree of dissociation `alpha` at a total pressure P byA. `K_(p)=(alpha^(3)P^(1//2))/((1+alpha)(2+alpha)^(1//2))`B. `K_(p)=(alpha^(3)P^(3//2))/((1-alpha)(2+alpha)^(1//2))`C. `K_(p)=(alpha^(3//2)P^(2))/((1-alpha)(2+alpha)^(1//2))`D. `K_(p)=(alpha^(3//2)P^(1//2))/((1-alpha)(2+alpha)^(1//2))` |
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Answer» Correct Answer - D Consider the equilibrium expression `{:(,H_(2)O(g),hArrH_(2)(g)+,(1)/(2)O_(2)(g)),("Inital moles",1,0,0),("Equilibrium moles",1-alpha,alpha,alpha//2):}` Total moles at equilibrium `=(1-alpha)+(alpha)+(alpha)/(2)` `=1+alpha//2` Note that is always convenient to consider 1 mol whenever we del with the degree of dissociation. At equilibrium, the partial pressure of gas can be calculated as follows: Partial pressure = (Mole fraction) x (Total pressure) `:. P_(H_(2)O)= (1-alpha)/(1+alpha//2)P` `= (2(1-alpha))/(2+alpha)P` `P_(H_(2))=(alpha)/(1+alpha//2)P=(2alpha)/(2+alpha)P` `P_(O_(2))=(alpha//2)/(1+alpha//2)P=(alpha)/(2+alpha)P` According to the law of chemical equilibrium, `K_(p)=(P_(H_(2))P_(O_(2))^(1//2))/(P_(H_(2)O))` `= (((2alpha)/(2+alpha)P)((alphaP)/(2+alpha))^(1//2))/((2(1-alpha))/(2+alpha)P)` `=(alpha)/(1-alpha)((alphaP)/(2+alpha))^(1//2)` `=(alpha^(3//2)P^(1//2))/((1-alpha)(2+alpha)^(1//2))` |
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