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22 January, 16:23

fills a 500. mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she determines that it contains 3.36atm of carbon monoxide gas, 0.96atm of water vapor and 0.24atm of hydrogen gas. The engineer then adds another 1.8atm of carbon monoxide, and allows the mixture to come to equilibrium again. Calculate the pressure of carbon dioxide after equilibrium is reached the second time. Round your answer to 2 significant digits.

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  1. 22 January, 16:48
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    The answer to the question is

    The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

    Explanation:

    To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

    At the first trial the mixture contains

    3.6 atm CO

    1.2 atm H₂O (g)

    Total pressure = 3.6+1.2 = 4.8 atm

    which gives

    3.36 atm CO

    0.96 atm H₂O (g)

    0.24 atm H₂ (g)

    That is

    CO+H₂O→CO (g) + H₂ (g)

    therefore the mixture contained

    0.24 atm CO₂ and the total pressure =

    3.36+0.96+0.24+0.24 = 4.8 atm

    when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

    At equilibrium we have 0.24*0.24 / (3.36*0.96) = 0.017857

    adding 1.8 atm CO gives 4.46 atm hence we have

    (0.24+x) (0.24+x) / (4.46-x) (0.96-x) = 0.017857

    which gives x = 0.031 atm or x = - 0.6183 atm

    Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
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