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March 25, 2017

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When CO2(g) (0.118 mol/L) and H2(g) (3.90 mol) in a 33.0 L reaction vessel at 980K are allowed to come to equilibrium the mixture contains 0.0853 mol/L of H20(g). What concentration (mol/L) of CO2(g) reacted?

CO2(g) + H2(g) <--> CO(g) + H2O(g)

  • Chemistry - ,

    If you will do the ICE chart I think you can see the answer.
    CO2 + H2 <==> H2O + CO

    I= initial concn:
    (CO2)=0.118 M
    (H2) = 3.90 M if the problem means mols/L or 3.90/33 = 0.118 M if it means mols in 33.0 L.
    (H2O) = 0
    (CO) = 0

    C = change in concn:
    (H2O) = +x
    (CO) = +x
    (CO2) = - x
    (H2) = -x

    E = equilibrium concn:
    (H2O) = 0.0853 from the problem.
    (CO) we know from the equation that this is also 0.0853.
    (CO2) = 0.118 - x but you know what x is. It is 0.0853.
    Check my work. Check my thinking.

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