anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
anonymous
  • anonymous
you got it??
anonymous
  • anonymous
@mastermind u der... i m in hurry ask ur doubt if u had in any step??

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anonymous
  • anonymous
uhm i got 30kJ :S
anonymous
  • anonymous
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
anonymous
  • anonymous
but y u mulitplie it by 4 to the last and by 2 to the second one
anonymous
  • anonymous
(3) 1/2N2(g) + 3/2O2(g) + 1/2H2(g)  HNO3(l) H0 = -174 kJ
anonymous
  • anonymous
so just realized the copy paste failed me...these are the coefficiets for the last reaction
anonymous
  • anonymous
sorry
anonymous
  • anonymous
is it right now?
anonymous
  • anonymous
yes .. i thought i made mistake :)
anonymous
  • anonymous
:) but you didn't reverse the first reaction or did you.
anonymous
  • anonymous
i did but in diff method in which no need to reverse but ye i forgot to multiply second reaction by 2 :P
anonymous
  • anonymous
okie dokie thank you :)
anonymous
  • anonymous
yw and bye..

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