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anonymous

  • 4 years ago

14. Calculate the enthalpy change of the following reaction between nitrogen gas and oxygen gas, given thermochemical equations (1), (2), and (3). 2N2(g) + 5O2(g)  2N2O5(g) (1) 2H2(g) + O2(g)  2H2O(l) H0 = -572 kJ (2) N2O5(g) + H2O(l)  2HNO3(l) H0 = -77 kJ (3) N2(g) + O2(g) + H2(g)  HNO3(l) H0 = -174 kJ

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  1. anonymous
    • 4 years ago
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    as u knw N2 is reactant side and it is 2 mole now luk in below reaction there is n2 but it is of one mole so multiply it by 2 =-174x2 -348 ok now O2 is common in all so neglect it u dont have to look coming to N2O5 its is product side of main reaction but in below reaction it is reatant side so we will change sign -x-77 +77 ok now by luking main reaction put enthaplies ok N2 + 5O2-------->N2O5 -348+77+(-572)=delta H (for calculating add all heats together ) -843 =delta H

  2. anonymous
    • 4 years ago
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    you got it??

  3. anonymous
    • 4 years ago
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    @mastermind u der... i m in hurry ask ur doubt if u had in any step??

  4. anonymous
    • 4 years ago
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    uhm i got 30kJ :S

  5. anonymous
    • 4 years ago
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    first i reversed the second reaction since N2)5 must be on the product side which mkes -77--->77) then i had to reverse the first reaction since water has to be on the reactant side to be cancelled out in the end, which makes -572 ---->572 then I multiplied the second reaction by 2 (77*2) then i multiplied the last reaction by 4 (-174*4) and at last i cancelled out the oxygen leaving 5o2 and then i cancelled out all the H2 as well as all the h2o. 572kJ (2)77kJ (4)-174kJ ________ 30kJ

  6. anonymous
    • 4 years ago
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    but y u mulitplie it by 4 to the last and by 2 to the second one

  7. anonymous
    • 4 years ago
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    (3) 1/2N2(g) + 3/2O2(g) + 1/2H2(g)  HNO3(l) H0 = -174 kJ

  8. anonymous
    • 4 years ago
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    so just realized the copy paste failed me...these are the coefficiets for the last reaction

  9. anonymous
    • 4 years ago
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    sorry

  10. anonymous
    • 4 years ago
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    is it right now?

  11. anonymous
    • 4 years ago
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    yes .. i thought i made mistake :)

  12. anonymous
    • 4 years ago
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    :) but you didn't reverse the first reaction or did you.

  13. anonymous
    • 4 years ago
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    i did but in diff method in which no need to reverse but ye i forgot to multiply second reaction by 2 :P

  14. anonymous
    • 4 years ago
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    okie dokie thank you :)

  15. anonymous
    • 4 years ago
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    yw and bye..

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