For the reaction
$SO subscript 2 left parenthesis straight g right parenthesis space plus space NO subscript 2 left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space space SO subscript 3 left parenthesis straight g right parenthesis space plus space NO left parenthesis straight g right parenthesis$ the partial pressure of SO₂, NO₂, SO₃ and NO at equilibrium are 0.5, 0.8, 0.7 and 1.2 bar respectively. Calculate K_p for the reaction.

Solution

Reversible reaction at equilibrium is

space SO subscript 2 left parenthesis straight g right parenthesis space plus NO subscript 2 left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space SO subscript 3 left parenthesis straight g right parenthesis space plus space NO left parenthesis straight g right parenthesis

Applying the law of chemical equilibrium

straight K subscript straight p space equals space fraction numerator straight p subscript SO subscript 3 end subscript cross times straight p subscript NO over denominator straight p subscript SO subscript 2 end subscript space cross times space straight p subscript NO subscript 2 end subscript end fraction

Substituting the values, we have,

straight K subscript straight p space equals space fraction numerator 0.7 space cross times space 1.2 over denominator 0.5 space cross times space 0.8 end fraction space equals space fraction numerator 0.84 over denominator 0.4 end fraction space equals space 2.1

Some More Questions From Equilibrium Chapter

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