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Question
One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation,
\[\ce{H2O (g) + CO (g) ⇌ H2 (g) + CO2 (g)}\]
Calculate the equilibrium constant for the reaction.
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Solution
The given reaction is:
| H2O(g) | + | CO(g) | ↔ | H2(g) | CO2(g) | |
| Initial conc. | `1/10`M | `1/10`M | 0 | 0 | ||
| At equilibrium | `(1 - 0.4)/10`M | `(1 - 0.4)/10`M | `0.4/10`M | `0.4/10`M | ||
| = 0.06 M | = 0.06 M | = 0.04 M | = 0.04 M |
Therefore, the equilibrium constant for the reaction,
`"K"_"c" = (["H"_2]["CO"_2])/(["H"_2"O"]["CO"])`
`= (0.04 xx 0.04)/(0.06 xx 0.06)`
= 0.444 (approximately)
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