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Standard XII

Chemistry

Derivation of Kp and Kc

One mole of ...

Question

One mole of $S O_{3}$ was placed in a 1 lit. reaction vessel at a constant temperature. The following equilibrium was established: $2 S O_{3} \Leftrightarrow 2 S O_{2} + O_{2}$ . At equilibrium, 0.6 moles of $S O_{2}$ were formed. The equilibrium constant of the reaction will be:

0.36

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0.45

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0.54

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0.675

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Solution

The correct option is A $0.675$
The molar concentrations are listed in the following table.

$S O_{3}$ $S O_{2}$ $O_{2}$
Initial 1 0 0
Change -2x 2x x
Equilibrium -2x 2x x
But at equilibrium, $[S O_{2}] = 2 x = 0.6$ or $x = 0.3$ .

Thus, the equilibrium concentrations of $S O_{3}, S O_{2}$ and $O_{2}$ are 0.4M, 0.6M and 0.3M respectively.

The expression for the equilibrium constant becomes $K_{c} = \frac{[S O_{2}]^{2} [O_{2}]}{[S O_{3}]^{2}} = \frac{(0.6)^{2} (0.3)}{(0.4)^{2}} = 0.675$ .

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	$S O_{3}$	$S O_{2}$	$O_{2}$
Initial	1	0	0
Change	-2x	2x	x
Equilibrium	-2x	2x	x

One mole of SO3 was placed in a 1 lit. reaction vessel at a constant temperature. The following equilibrium was established: 2SO3⇔2SO2+O2. At equilibrium, 0.6 moles of SO2 were formed. The equilibrium constant of the reaction will be:

One mole of $S O_{3}$ was placed in a 1 lit. reaction vessel at a constant temperature. The following equilibrium was established: $2 S O_{3} \Leftrightarrow 2 S O_{2} + O_{2}$ . At equilibrium, 0.6 moles of $S O_{2}$ were formed. The equilibrium constant of the reaction will be: