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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

One mole of ...

Question

One mole of $S O_{3}$ was placed in a litre reaction vessel at a certain temperature , the following equilibrium was established, $2 S O_{3} ⇌ 2 S O_{2} + O_{2}$ . 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 D 0.675
Solution:- (D) $0.675$
Given that at equilibrium, $0.6$ moles of $S O_{2}$ were formed.
$Initially At equilibrium 2 S O_{3} 1 (1 - 0.6) ⇌ 2 S O_{2} 0 0.6 + O_{2} 0 0.3$
Equilibrium constant $(K) = \frac{{[S O_{2}]}^{2} [O_{2}]}{{[S O_{3}]}^{2}} = \frac{{(0.6)}^{2} \times 0.3}{{(0.4)}^{2}} = 0.675$
Hence t equilibrium constant for the given the reaction will be $0.675$ .

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One mole of SO3 was placed in a litre reaction vessel at a certain 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:

The correct option is D 0.675Solution:- (D) 0.675Given that at equilibrium, 0.6 moles of SO2 were formed.InitiallyAt equilibrium2SO31(1−0.6)⇌2SO200.6+O200.3Equilibrium constant (K)=[SO2]2[O2][SO3]2=(0.6)2×0.3(0.4)2=0.675Hence t equilibrium constant for the given the reaction will be 0.675.

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