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-   -   When hydrogen gas and iodine gas are mixed in a vessel what happens pleas help.. (https://www.askmehelpdesk.com/showthread.php?t=551813)

  • Feb 6, 2011, 06:51 AM
    Spud93
    When hydrogen gas and iodine gas are mixed in a vessel what happens pleas help..?
    When hydrogen gas and iodine gas are mixed in a vessel at constant
    temperature, a reaction occurs which, after a certain amount of time, reaches
    equilibrium giving a mixture of the reactants and a product (hydrogen iodide).
    The balanced chemical equation for this reaction is:
    H2(g)+I2(g)= 2HI(g) (2)

    (a)
    (I) Write down the equilibrium expression for Reaction 2.
    (ii) For the equilibrium reaction between hydrogen and iodine, apply
    Le Chatelier's principle to explain what effect increasing and
    decreasing the pressure will have on the equilibrium yield of hydrogen
    iodide in Reaction 2. (We expect that you will be able to answer part
    (a) (ii) in about 100 words.)
    (b) Table 2 gives the results of several experiments for Reaction 2 where three
    different mixtures of H2
    and I2
    gases were allowed to react and reach
    equilibrium at a temperature of 427 °C. From the data in Table 2, and
    using your equilibrium expression from part (a) (I), calculate the average
    value of the equilibrium constant, K, for Reaction 2 at 427 °C. Give all
    your values for K, including the final average value of K, to one decimal
    place.
    Table 2 Concentrations of H2, I2 and HI gases at equilibrium for three
    different mixtures at 427 °C
    Experiment::::[H2]/mol dm−3::::[I2]/mol dm−3::::[HI]/mol dm−3
    1 ::::4.80 × 10−4 ::::4.80 × 10−4 ::::3.53 × 10−3
    2 ::::1.14 × 10−3 ::::1.14 × 10−3 ::::8.41 × 10−3
    3 ::::4.56 × 10−3 ::::7.40 × 10−4 ::::1.35 × 10−2
    (c) Table 3 gives values for the equilibrium constant at several temperatures.
    The missing value of K at 427 °C is one you have determined in part (b).
    (I) From the data in Table 3, how does the value of the equilibrium
    constant, K, change with respect to temperature? Does the value you
    determined in part (b) fit with this trend? (A couple of sentences) You
    do not need to copy out Table 3 in your answer.
    (ii) Would you expect this reaction to be exothermic or endothermic? Give
    brief reasoning of how you have arrived at your answer in light of the
    data given in Table 3. (Three or four sentences)
    Table 3 The equilibrium constant, K, for Reaction 2 as a function of
    temperature

    T/^oC K
    25 794.4
    227 128.3
    327 81.1
    427
    527 43.6
    627 35.7
  • Feb 6, 2011, 10:26 AM
    Unknown008

    It would be good if you post your attempt at the question. Part (a) can normally be found in notes or in your textbook, part (b) is just a matter of plugging in values and getting an average. Part (c) requires you to analyse the data you got and make deductions.
  • Feb 7, 2011, 05:05 AM
    Spud93
    4a.I) K = (HI)^2 / (H2) (I2)
    ii) Increasing the pressure will favor the side with fewer mols and the and the opposite for decreasing pressure. However, since we have two mols of reactants and two mols of product changing the pressure will have no effect on the equilibrium position (this is a shortened version of my answer) as for b. an c. I'm stuck :(
  • Feb 7, 2011, 06:14 AM
    Unknown008

    a) (i) Actually, what I would have put is:



    since we are dealing with gases and when we do so, the equation is in terms of partial pressure.

    (ii) is okay, though I'd have liked to see your full answer.

    b)

    Use the first row and calculate

    Here's the first one:



    Can you do the rest now?

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