Question

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

Answer 1

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

Answer 2

you got it??

Answer 3

@mastermind u der... i m in hurry ask ur doubt if u had in any step??

Answer 4

uhm i got 30kJ :S

Answer 5

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

Answer 6

but y u mulitplie it by 4 to the last and by 2 to the second one

Answer 7

(3) 1/2N2(g) + 3/2O2(g) + 1/2H2(g)  HNO3(l) H0 = -174 kJ

Answer 8

so just realized the copy paste failed me...these are the coefficiets for the last reaction

Answer 9

sorry

Answer 10

is it right now?

Answer 11

yes .. i thought i made mistake :)

Answer 12

:) but you didn't reverse the first reaction or did you.

Answer 13

i did but in diff method in which no need to reverse but ye i forgot to multiply second reaction by 2 :P

Answer 14

okie dokie thank you :)

Answer 15

yw and bye..