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Question

 Equilibrium constants are given for the following two equilibria. 
$$(i) A_2(g)+ B_2 (g)\rightleftharpoons 2AB (g);$$          $$K=2 \times 10^{-4}$$
$$(ii) 2AB(g)+ C_2(g) \rightleftharpoons 2ABC (g);$$  $$K= 2\times 10^{-2}\, L \, mol^{-1}$$

Calculate the equilibrium constant for the following equilibrium. 
$$ABC (g)\rightleftharpoons \dfrac {1}{2}A_2 (g)+\dfrac {1}{2}B_2(g)+\dfrac {1}{2}C_2 (g)$$


A
500 M1/2
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B
4×108 M1/2
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C
500M1/2
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D
200M1/2
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E
500M1/2
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Solution

The correct option is E $$500\, M^{1/2}$$
$$A_2(g)+B_2(g)\rightleftharpoons 2AB(g)$$

$$K_1 = 2\times 10^{-4}$$

$$K_1=\dfrac {[AB]^2}{[A_2][B_2]}$$    .... (i)

$$2AB (g)+ C_2(g)\rightleftharpoons 2ABC(g)$$

$$K_2 = 2 \times 10^{-2}L/mol$$

So, $$K_2 =\dfrac {[ABC]^2}{[AB]^2[C_2]}$$  .... (ii)

$$ABC \rightleftharpoons \dfrac {1}{2}A_2(g)+\dfrac {1}{2}B_2 (g)+\dfrac {1}{2}C_2(g)$$

So, $$K_3 =\dfrac {[A_2]^{1/2}[B]^{1/2}[C_2]^{1/2}}{[ABC]}$$

Eq. (i) and Eq. (ii) are added, reversed and divided by 2 then we get Eq. (iii).
Therefore, $$K_3 =\dfrac {1}{\sqrt{K_1 \times K_2}}$$

$$=\dfrac {1}{\sqrt{2\times 10^{-4}\times 2 \times 10^{-2}}}$$

$$=\dfrac {1000}{2} = 500 \, M^{1/2}$$

Chemistry

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