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Intro Chemistry: Does this look ok?

  1. Mar 9, 2006 #1
    Our professor gave us three questions for our exam tomorrow, I was wondering if anyone would check my work for me. I just want to make sure I am doing this correctly. Thanks in advance.

    15) Which of the following species has the largest atomic size:
    (a) [tex] S^{2-} [/tex]
    (b) [tex] Cl^{-} [/tex]
    (c) [tex] Ca^{2+} [/tex]
    (d) [tex] Ar [/tex]

    15.A: They are an isoelectric series since they have the same number of electrons, thus the largest radius is determined by the isotope with the smallest atomic number, which is:
    (a) [tex] S^{2-} [/tex]

    ----------------------------------------------------
    16) Given the following reactions:
    [tex] N_{2(g)} + 2O_{2(g)} \longrightarrow 2NO_{2(g)} \,\,\, \Delta H_1 = 66.4kJ [/tex]
    [tex] 2NO_{(g)} + O_{2(g)} \longrightarrow 2NO_{2(g)} \,\,\,\Delta H_2 = -114.2kJ [/tex]

    What is [itex] \Delta H_{rxn} [/itex] for the following process?
    [tex] N_{2(g)} + O_{2(g)} \longrightarrow 2NO(g) [/tex]

    (a) 180.6kJ
    (b) -47.8kJ
    (c) 47.8kJ
    (d) 90.3kJ
    (e) -180.6kJ

    16.A:
    [tex] N_{2(g)} + 2O_{2(g)} \longrightarrow 2NO_{2(g)} + \Delta H_1 = 66.4kJ [/tex]
    [tex] 2NO_{(g)} + O_{2(g)} \longrightarrow 2NO_{2(g)} + \Delta H_2 = -114.2kJ [/tex]
    ---
    [tex] O_{2(g)} \longrightarrow -2NO_{(g)} + 2NO_{2(g)} + \Delta H_2 = -114.2kJ [/tex]
    [tex] -O_{2(g)} \longrightarrow 2NO_{(g)} - 2NO_{2(g)} - \Delta H_2 = -114.2kJ [/tex]

    [tex] N_{2(g)} + O_{2(g)} \longrightarrow 2NO_{(g)} + \Delta H_1 - \Delta H_2 [/tex]

    Thus: [itex] \Delta H_1 - \Delta H_2 = (66.4+114.2)kJ = 180.6kJ [/itex]

    ----------------------------------------------------
    19) All of the following are characteristics of Alkali metals except
    (a) They are very reactive.
    (b) They have low densities and low melting points
    (c) They have high thermal and electrical conductivities
    (d) They have a fully filled VE shell of electrons
    (e) They react w/ [itex] O_2 [/itex] to form oxides, peroxides, and superoxides.

    19.A:
    a, b, c, e are all true.
    Thus the answer is (d)
     
    Last edited: Mar 9, 2006
  2. jcsd
  3. Mar 10, 2006 #2
    Great going ... Everything seems right to me
     
  4. Mar 10, 2006 #3
    Cool. I really appreciate you looking over it :)

    Have a good one. Now... I just have to go take the test without notes :(
     
  5. Mar 10, 2006 #4
    Well... they are not really isotopes.
    Remember isotopes are atoms of same element but with different amount of neutrons.


    All your answers seem right anyway.
     
  6. Mar 10, 2006 #5
    Could someone explain what Frogpad has done in the exercise 16? I would like to understand. Thank you!
     
  7. Mar 10, 2006 #6
    Well im not sure about the technique he used.... it is kind of different, and harder to understand. All you got to do in this problem is reverse the second equation and change the sign of the Enthalpy of that equation.
    Then when you add the first and modified second equation, some of the reactants and products of any of the equations cancel out leaving the desired equation which was
    N2 + O2 ---> 2NO
    and you add the enthalpy values to get the enthalpy of that desired equation.


    If this is the first time you have seen these kind of problems, then my repsonse may be of no use to you, so use your textbook or search online for tuts.
     
  8. Mar 10, 2006 #7
    Haxx0rm4ster, thank you, but you are just explaining what he has done and not why he has done what he did.
     
  9. Mar 10, 2006 #8
    Last edited: Mar 10, 2006
  10. Mar 10, 2006 #9
    Woops, I meant to say Ions, not isotopes.

    Haxx0rm4ster: Are you asking for the method that I used? Actually, I'll just answer you assuming this is the question you are asking.

    In my chem book, they talk about Hess's law and rules that can be used to perform Hess's law. I'll give you an example:

    Say you have:
    [tex] 2NO_{(g)} + O_{2(g)} \longrightarrow 2NO_{2(g)} \,\,\,\,\, \Delta H = -114.2kJ [/tex]

    The book says that you can FLIP the order of the reactants and the products which would be:
    [tex] 2NO_{2(g)} \longrightarrow 2NO_{(g)} + O_{2(g)} [/tex]

    But, when you flip them you have to change the sign of [itex] \Delta H [/tex] so the flip would become:
    [tex] 2NO_{2(g)} \longrightarrow 2NO_{(g)} + O_{2(g)} \,\,\,\,\, \Delta H = 114.2kJ [/tex]

    Now, yeah... that's cool and all. But why does that work? Well our T.A. showed us a "method" or maybe it should be called a "trick", but it makes MUCH more sense to me. Instead of tacking the [itex] \Delta H [/itex] on the end of the equation, instead include it as an algebraic quantity.

    So instead we would have:
    [tex] 2NO_{(g)} + O_{2(g)} \longrightarrow 2NO_{2(g)} + \Delta H [/tex]
    In this form, some algebraic rules apply (I don't know the extent of which rules are applicable... I know addition, subtraction, and multiplying by a constant hold)

    Now lets say you want to "flip" the equation. You can now use algebre, so you would do it like this:

    [tex] O_{2(g)} \longrightarrow -2NO_{(g)} + 2NO_{2(g)} + \Delta H [/tex]
    [tex] \longrightarrow -O_{2(g)} -2NO_{(g)} + 2NO_{2(g)} + \Delta H [/tex]
    [tex] - 2NO_{2(g)} \longrightarrow -O_{2(g)} -2NO_{(g)} + \Delta H [/tex]

    Now we need to multiply by a constant:
    [tex](-1) - 2NO_{2(g)} \longrightarrow (-1)-O_{2(g)} -2NO_{(g)} + \Delta H = -114.2kJ [/tex]
    [tex] 2NO_{2(g)} \longrightarrow O_{2(g)} + 2NO_{(g)} - \Delta H[/tex]

    Now if we take [itex] \Delta H [/itex] out we have:
    [tex] -\Delta H = -114.2kJ [/tex]
    [tex] \Delta H = 114.2kJ [/tex]

    Does that answer your question?
     
    Last edited: Mar 10, 2006
  11. Mar 10, 2006 #10
    lol, well I wasnt really asking, but there is another way which is a LOT more simple.


    Easy way I learned:
    http://en.wikipedia.org/wiki/Hess's_Law


    except they use 4 equations to find 1 ... but still not very hard to understand.
     
  12. Mar 10, 2006 #11
    lol...sorry about that. I meant to copy and paste "PPonte" as the name (to the person I addressed) :)
     
    Last edited: Mar 10, 2006
  13. Mar 11, 2006 #12
    Frogpad, thank you.
    Simple, good and neat explanation.

    Haxx0rm4ster, thank you.
    I didn't know that this was a law. If I knew I would have searched and didn't have to bore you.
     
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