Chemical EquilibriumHard
Question
An amount of 1 mole each of A and D is introduced in 1 L container. Simultaneously the following two equilibria are established. A$\rightleftharpoons$B + C: KC = 106 M and B + D$\rightleftharpoons$A: KC = 10−6 M−1
Options
A.10−6 M
B.10−3 M
C.10−12 M
D.10−4 M
Solution
A $\rightleftharpoons$ B + C; K1 = 106
Initial moles 1 0 0
Equilibrium moles 1 – x + y x – y x
B + D $\rightleftharpoons$ A; K2 = 10–6
x 1 1
Equilibrium moles x – y 1 – y 1 + y – x
As K1 >> 1, we may assume x ≈ 1
Now, $K_{2} = \frac{(1 + y - x)}{(x - y)(1 - y)} \approx \frac{y}{(1 - y).(1 - y)}$
As K2 << 1, we may assume y << 1
$K_{2} = \frac{y}{(1 - y)(1 - y)} \simeq y $$$\therefore\lbrack A\rbrack = 1 - x + y \simeq y = 10^{- 6}\text{ M}$$
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