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Use the following data to calculate a value for the Xe-F bond enthalpy in XeF4 watch

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    Use the following data to calculate a value for the Xe-F bond enthalpy in XeF4
    Xe(g) + 2F2(g) -> XeF4(g) delta H = -252 kJ/mol
    F2(g) -> 2F(g) delta H = +158 kJ/mol

    Is this a hess's law question or a mean enthalpy question (bonds broken - bonds formed). Not sure how to do this, any help is greatly appreciated thanks!
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    (Original post by Tim73)
    Use the following data to calculate a value for the Xe-F bond enthalpy in XeF4
    Xe(g) + 2F2(g) -> XeF4(g) delta H = -252 kJ/mol
    F2(g) -> 2F(g) delta H = +158 kJ/mol

    Is this a hess's law question or a mean enthalpy question (bonds broken - bonds formed). Not sure how to do this, any help is greatly appreciated thanks!
    Bit of both ...

    At the end of the day all enthalpy questions boil down to Hess' law ... which is just another way of stating the law of conservation of energy.

    For this question you first have to write out an equation for the average bond enthalpy of the Xe-F bond.

    XeF4(g) --> Xe(g) + 4F(g) ........... ΔH = 4 x bond enthalpy Xe-F

    Now use the equations given to construct a cycle with this equation.
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    (Original post by charco)
    Bit of both ...

    At the end of the day all enthalpy questions boil down to Hess' law ... which is just another way of stating the law of conservation of energy.

    For this question you first have to write out an equation for the average bond enthalpy of the Xe-F bond.

    XeF4(g) --> Xe(g) + 4F(g) ........... ΔH = 4 x bond enthalpy Xe-F

    Now use the equations given to construct a cycle with this equation.
    Thanks, so is this enthalpy of formation right? So it will be the sum of enthalpy change of the products - the sum of the enthalpy change of the reactants?
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    (Original post by Tim73)
    Thanks, so is this enthalpy of formation right? So it will be the sum of enthalpy change of the products - the sum of the enthalpy change of the reactants?
    No, it's a mixture of enthalpy changes ...
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    i got +142 kj/mol
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    (Original post by richpanda)
    i got +142 kj/mol
    Would you please be able to explain how? I'm still not sure I understand this. Thanks
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    (Original post by Tim73)
    Would you please be able to explain how? I'm still not sure I understand this. Thanks
    Check this out ...

    Name:  Xe-F.jpg
Views: 1455
Size:  14.2 KB
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    (Original post by charco)
    Check this out ...

    Name:  Xe-F.jpg
Views: 1455
Size:  14.2 KB
    Thanks a lot for your help. Think I get it now.
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    The enthalpy in the Q is 252, however in workinh it is 242. Just a working error?
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    (Original post by charco)
    Check this out ...

    Name:  Xe-F.jpg
Views: 1455
Size:  14.2 KB
    Why did we make the third equation??
    I am so confused, thank you!
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    (Original post by Minawembly)
    Why did we make the third equation??
    I am so confused, thank you!
    The third equation is the definition of bond enthalpy (x 4 in this case)

    Bond enthalpy = "the energy required to break 1 mole of bonds in gaseous molecules measured over a range of similar structures"

    Xe-F does not exist as a molecule so:

    XeF4(g) --> Xe(g) + 4F(g) is equal to four times the Xe-F bond enthalpy term.
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    (Original post by charco)
    The third equation is the definition of bond enthalpy (x 4 in this case)

    Bond enthalpy = "the energy required to break 1 mole of bonds in gaseous molecules measured over a range of similar structures"

    Xe-F does not exist as a molecule so:

    XeF4(g) --> Xe(g) + 4F(g) is equal to four times the Xe-F bond enthalpy term.
    Thank you ! Makes so much more sense now!
    What does the delta H sign stand for ?
    Is it the energy needed to form those products?
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    (Original post by Minawembly)
    Why did we make the third equation??
    I am so confused, thank you!
    Why do the arrows go down? And how do we know when to do that ?
 
 
 
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