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Partial Pressures Problem

  1. Oct 1, 2008 #1
    C6H12O6 (glucose) +6O2 ---> 6CO2 + 6H2O

    If this reaction is carried out in an expandable container at 35degrees C and 780 torr, what is the partial pressure of each gas when the reaction is 50% complete (9.0 g of glucose remains)?

    I know that 35 degrees C = 308.15k and that 780 torr = 1.03 atm. I also know that at 35degrees C, the vapor pressure of water is always 42.2 torr (0.056 atm).
    If 9.0g of glucose remain, then there are 0.05 mol of glucose, 0.3 mol of CO2, 0.3 mol of O2, and 0.3 mol of H2O

    I subtracted 0.056 atm from 1.03 atm to get .974 atm and used the equatoin Partial Pressure of X = (Molar Ratio of X) x Total Pressure. (Molar Ratio = Mol X/ Total Mols in Sample). However, this did not produce the answer in the back of the book, which reads that the partial pressures of CO2 and H2O are both 3.7 x 10^2 torr.
  2. jcsd
  3. Oct 2, 2008 #2


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    Staff: Mentor

    Most likely what they mean is that CO2 and O2 have both partial pressures of 370 torr. 370 torr CO2 + 370 torr O2 + 42 H2O gives around 780, which fits.
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