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Sublimation and Pressure

  1. Nov 17, 2013 #1


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    Gold Member

    1. The problem statement, all variables and given/known data

    When fluorine and solid iodine are heated at 550. K, the iodine completely sublimes and gaseous iodine heptafluoride forms. If 350. torr of fluorine gas and 2.50 g of solid iodine are put into a 2.50 L container at 250. K and the container is heated to 550. K, what is the final pressure in the flask?

    2. Relevant equations

    5F2 (g) + I2 (s) --> 2IF5(g)

    3. The attempt at a solution

    We have 350/760 atm of fluorine gas. Using the ideal gas law, P = nRT/V, we find taht we have 0.056 moles of F.

    We also have 0.009 moles of solid iodine (2.5 grams / twice the molecular mass of elemental iodine).

    This yields 0.0197 moles of product. Iodine is limiting and fluorine is in excess. The final pressure, by Dalton's Law of Partial Pressures, is the pressure of the product and the pressure of the remaining reactants.

    We have 0.0197 * (5/2) moles of fluorine used. This means we have 0.056 minus 0.0475 moles of fluorine remaining. This means there are 0.085 moles of fluorine remaining. The pressure of this remaining fluorine is:

    P = nRT/V = 0.085(0.0821)(550)/2.5 = 0.15 atm.

    The pressure of the product is also nRT/V = 0.0197 * 0.0821 * 550 / 2.5 = 0.355 atm.

    Summing these two pressures, we get 0.50 atm of pressure. However, this isn't an answer choice. The closest choice is 0.534 atm, leading me to believe there is an error in my calculations, and not a rounding one at that. Is there any flaw in my line of reasoning?
  2. jcsd
  3. Nov 17, 2013 #2


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    Staff: Mentor

    What is the formula of the iodine heptafluoride?
  4. Nov 17, 2013 #3


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    Gold Member

    Oh snap. Hept ... that's prefix for 7. I confused that with pentafluoride. Thanks for the catch!
  5. Nov 19, 2013 #4


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    Okay, new balanced equation:

    7F2 + I2 --> 2IF7.

    New process.

    We have to convert everything to atmospheres of pressure at 550 K to solve this problem. This is the most convenient way. My method doesn't work because it doesn't account for the temperature change and how that might affect the Fluorine in terms of pressure. I can't use pressure of F at the initial temperature and arrive at an amount of Fluorine; it'll change when the temperature changes.

    1) Find the atmospheres of I2 at 550 K. P = nRT/V. P = 0.177 atm.

    2) Find the atmospheres of F2 at 550 K. P / T = P / T. Convert torr to atm (350/760) and temperatures are in Kelvins (250 and 550 K). Solve for the new pressure of F2.

    3) Now we can set up this classic problem (classic for my teacher; he often gives us atmospheres of pressure and has us to a stoichiometry problem; this just takes it a step further by withholding the proper pressure at the proper temperature). Use atmospheres of pressure for each reactant and find which one will result in the least amount (pressure) of product. Total pressure = pressure of product + pressure of left over reactants.
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