One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 60% of water (by mass) reacts with CO according to the equation,

H2O(g)  +  CO(g) ⇌ H2(g)   +  CO2(g)

Calculate the equilibrium constant for the reaction.

Asked by Pragya Singh | 11 months ago |  68

##### Solution :-

The given reaction is:

H2O(g)  +  CO(g) ⇌ H2(g)   +  CO2(g)

 Compound H20 CO H2 CO2 Initial Conc. 0.1M 0.1M 0 0 Equilibrium Conc. 0.06M 0.06M 0.04M 0.04M

Therefore, the equilibrium constant for the reaction,

Kc = $$\dfrac{ ([H_2][CO_2])}{([H_2O][CO]) }$$

$$\dfrac{ (0.4\times0.4)}{(0.6\times 0.6)}$$ = 0.444

Answered by Abhisek | 11 months ago

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