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The dissociation equilibrium of a gas `AB_(2)` can be represented as : `2AB_(2) hArr 2AB(g) +B_(2)(g)` The degree of dissocaition x is very small as compared to 1. The expression which relates the degree of dissociation (x) with equilibrium constant `(K_(p))` and total pressure (p) is :A. `(2 K_(p) //P)^(1//2)`B. `(K_(p)//P)`C. `(2K_(p)//P)`D. `(2 K_(p)//P)^(1//3)` |
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Answer» Correct Answer - D `2AB_(2) (g) hArr 2AB (g) + B_(2)(g)` `"Initial mol:"" "2" "0" "0` `"Eqm mol:"" "2(1-x)" "2x" "x` No of moles at equilibrium. `=2(1 -x) + 2x +x =(2 + x) " mol "` `K_(p) = (p^(2) AB xx pB)/(p^(2) AB_(2))` `=(((2x)/(2+x))^(2) xx ((x)/(2+x)))/(((2(1-x))/((2+x)))^(2))=((4x^(3))/((2+x))xxP)/(4(1-x)^(2))` `K_(p) =((4x^(3) xx P)/(2))/(4) [2+xx~~2 " and " 1-x~~1]` `K_(p) =(x^(3)p)/(2) " or " x^(3) =(2Kp)/(P)` `x= ((2Kp)/(P))^(1//3)` |
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