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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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
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
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
you got it??
@mastermind u der... i m in hurry ask ur doubt if u had in any step??
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uhm i got 30kJ :S
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.
but y u mulitplie it by 4 to the last and by 2 to the second one