Question

One mole of H₂O and one mole of CO are taken in a 10 L vessel and heated to 725K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation

$straight H subscript 2 straight O left parenthesis straight g right parenthesis space plus space CO left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space space straight H subscript 2 left parenthesis straight g right parenthesis space plus space CO subscript 2 left parenthesis straight g right parenthesis$

Calculate the equilibrium constant for the reaction.

Solution

The given reversible reaction is

straight H subscript 2 straight O left parenthesis straight g right parenthesis space plus space CO space space rightwards harpoon over leftwards harpoon space straight H subscript 2 left parenthesis straight g right parenthesis space plus space CO subscript 2 left parenthesis straight g right parenthesis

At equilibrium, the concentration of various species are

open square brackets straight H subscript 2 straight O close square brackets space equals space fraction numerator 1 minus 0.40 over denominator 10 end fraction mol space straight L to the power of negative 1 end exponent space equals space 0.06 space mol space straight L to the power of negative 1 end exponent open square brackets CO close square brackets space equals space 0.06 space mol space straight L to the power of negative 1 end exponent open square brackets straight H subscript 2 close square brackets space equals space fraction numerator 0.4 over denominator 10 end fraction mol space straight L to the power of negative 1 end exponent space equals space 0.04 space mol space straight L to the power of negative 1 end exponent open square brackets CO subscript 2 close square brackets space equals space 0.04 space mol space straight L to the power of negative 1 end exponent

Applying the law of chemical equilibrium

straight K space equals space fraction numerator open square brackets straight H subscript 2 left parenthesis straight g right parenthesis close square brackets space open square brackets CO subscript 2 left parenthesis straight g right parenthesis close square brackets over denominator space open square brackets straight H subscript 2 straight O left parenthesis straight g right parenthesis close square brackets space open square brackets CO left parenthesis straight g right parenthesis close square brackets end fraction space space space space space... left parenthesis 1 right parenthesis

Putting the values in expression (1), we have,

straight K space equals space fraction numerator 0.04 space cross times space 0.04 over denominator 0.06 space cross times space 0.06 space end fraction space space equals space 0.444

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