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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
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as 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 x77 +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
 4 years ago
Best ResponseYou've already chosen the best response.0@mastermind u der... i m in hurry ask ur doubt if u had in any step??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first 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
 4 years ago
Best ResponseYou've already chosen the best response.0but y u mulitplie it by 4 to the last and by 2 to the second one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(3) 1/2N2(g) + 3/2O2(g) + 1/2H2(g) HNO3(l) H0 = 174 kJ

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so just realized the copy paste failed me...these are the coefficiets for the last reaction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes .. i thought i made mistake :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0:) but you didn't reverse the first reaction or did you.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i did but in diff method in which no need to reverse but ye i forgot to multiply second reaction by 2 :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okie dokie thank you :)
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