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    Help on question 2 pls

    EricPiphany samb1234
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    (Original post by thefatone)
    Help on question 2 pls

    EricPiphany samb1234
    \frac{1}{2}N_2(g)+\frac{3}{2}H_2  (g) \rightarrow NH_3(g)

    You know \boxed{\Delta H_b \ \text{of }H_2}, \boxed{\Delta H_b \  \text{of } \  N_2} and \boxed{\Delta H_f \  \text{for }  NH_3}
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    (Original post by Kvothe the arcane)
    \frac{1}{2}N_2(g)+\frac{3}{2}H_2  (g) \Rightarrow NH_3(g)
    multiple don't make a difference right? so i put a 3 in from of the H2 and a 2 in from of the ammonia

    so what did you work out the average bond enthalpy as?
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    (Original post by thefatone)
    Help on question 2 pls

    EricPiphany samb1234
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    (Original post by EricPiphany)
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    sure? the answers tell me +390.8 kJmol-1 but i get the same answer as you.....
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    (Original post by thefatone)
    sure? the answers tell me +390.8 kJmol-1 but i get the same answer as you.....
    I get 390.83 \  kJmol^{-1}

    Your mistake is that enthalpy change of formation is when one mole of the product is formed.
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    (Original post by Kvothe the arcane)
    I get 390.83

    Your mistake is that enthalpy change of formation is when one mole of the product is formed.
    You're correct.
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    (Original post by thefatone)
    multiple don't make a difference right? so i put a 3 in from of the H2 and a 2 in from of the ammonia

    so what did you work out the average bond enthalpy as?
    No. But then you would have to multiply -46 by 2 as 2 moles of NH3 are being formed. Your answer isn't that far off as you're dealing with large quantities and 46 is small in comparison.
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    (Original post by Kvothe the arcane)
    No. But then you would have to multiply -46 by 2 as 2 moles of NH3 are being formed. Your answer isn't that far off as you're dealing with large quantities and 46 is small in comparison.

    ah right i forgot thanks a ton guys

    (Original post by EricPiphany)
    You're correct.
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    I guess I'm too late
 
 
 
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