One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO 2 . What are the equilibrium partial pressures of CO and CO 2 at 1050 K if the initial pressure are: - Wired Faculty

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One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.
$FeO left parenthesis straight s right parenthesis space plus space CO left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space space Fe left parenthesis straight s right parenthesis space plus space CO subscript 2 left parenthesis straight g right parenthesis semicolon space space straight K subscript straight p space equals space 0.265 space atm space at space 1050 straight K$
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial pressure are: $straight p subscript co space equals space 1.4 space atm space and space straight p subscript co subscript 2 end subscript space equals 0.80 space atm$

Solution

$space space space space space space space space space space space space FeO left parenthesis straight s right parenthesis space plus CO left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon Fe left parenthesis straight s right parenthesis space plus CO subscript 2 left parenthesis straight g right parenthesis In. space pre. space space minus space space space space space space space 1.4 space atm space space space space space space space minus space space space space space space space 0.80 space atm space space space space space straight Q subscript straight p space equals end subscript space straight p subscript CO subscript 2 end subscript over straight p subscript CO equals fraction numerator 0.80 over denominator 14 end fraction equals 0.571 Since space straight Q subscript straight p greater than straight K subscript straight p comma space end subscript the space net space reaction space will space proceed space in space the space backward space direction. space Therefore comma space pressure space of space CO subscript 2 space will space derease space and space that space of space CO space will space increase space to space atttain space equilibrium. Hence space comma space if space straight p space is space the space decrease space in space pressure space of space CO subscript 2 comma space then space increase space in space pressure space of space CO equals straight p. therefore comma At space equilibrium comma straight p subscript CO subscript 2 end subscript space equals left parenthesis 0.80 minus straight p right parenthesis atm comma straight p subscript CO space equals left parenthesis 1.4 space plus straight p right parenthesis atm straight K subscript straight p space equals straight p subscript CO subscript 2 end subscript over straight p subscript CO 0.265 equals fraction numerator 0.80 minus straight p over denominator 14 plus straight p end fraction 0.265 left parenthesis 1.4 plus straight p right parenthesis space equals 0.80 minus straight p or 0.371 space plus 0.265 straight p space equals 0.80 minus straight p 1.265 straight p space equals 0.429 space space or space space straight p space equals 0.339 atm straight p subscript CO equals space 1.4 space plus 0.339 space atm equals 1.739 space atm straight p subscript CO subscript 2 end subscript space equals 0.80 minus 0.339 space atm space equals space 0.461 space atm$

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Equilibrium

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