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Q.

Consider the following reaction taking place in 1 L capacity container at 300 K
$A + B ⇌ C + D$
If one mole each of A and B are present initially and at equilibrium 0.7 mol of C is formed, then equilibrium constant (K_C) for the reaction is

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a

1.2

b

5.4

c

6.2

d

9.7

answer is C.

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Detailed Solution

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A+B⇌C+D

Initial	1 mole	1 mole	0	0
at equilibrium	1-x	1-x	x	x

x is degree of dissociation
given x=0.7
[A]=1-0.7=0.3
[" " B]=1-0.7=0.3
[C]=0.7
[D]=0.7
$K_{C} = \frac{[C] [D]}{[A] [B]} = \frac{0.7 \times 0.7}{0.3 \times 0.3} = \frac{49}{9} = 5.4$

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