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Consider the following equilibrium reaction at T K in a 1L flask. (I) `A(g)iffB(g)+C(g), K_(C_(1))=3` (II) `B(g)iff D(g)+C(g), K_(C_(2))=?` If initially 2 moles of A(g) are taken and allowed attain equilibrium, concentration of C(g) was formed to be 3M. What is the value of `K_(C )` of second reaction ?A. `(200)/(3)`B. `(266)/(9)`C. 12D. `(3)/(200)` |
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Answer» Correct Answer - C `" "A(g) iff B(g) + C(g), K_(C_(1)) = 3` At equilibrium mole `" "2-x " (x-y) (x+y)"` `" "B(g) iff D(g) + C(g), K_(C_(2)) = ?` At equilibrium mole `" "(x-y) " y (x+y)"` `K_(C_(1)) = 3 = ((x-y)(x+y))/(2-x)" ".....(1)` `K_(C_(2)) = (y(x+y))/(x-y) " ".....(2)` `x + y = 3 " "....(3)` 2 - x = x - y 2 = 2x - y x + y = 3 `x = 5//3, y = 3 - (5)/(3) = (4)/(3)` `K_(C_(2)) = ((4//3)(3))/((5//3-4//3)))=12` |
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