111 Calculate Hno for the reaction N2H4 l O2 g N2 g 2 H2O
Solution
Given reaction is
 
 N2H4(l) + O2(g) ---> N2(g) + 2H2O(l)
 
 consider reaction for Eliminating NH3
 
 2(NH3) + 3(N2O) ---> 4(N2) + 3(H2O)    H -1010.0 kJ
 +
 N2H4 + H2O ---> 2(NH3) + 1/2(O2) H +143 kJ (reaction reverse H become +ve)
 --------------------------------------------------------------------
  N2H4 + 3(N2O) ---> 4(N2) + 2(H2O) + 1/2(O2)     H -867 kJ   -----------(A)
 
 After the addition, there is 2(NH3) on both sides, we cancel it.
 For H2O, there is 1 on the left and 3 on the right. Canceling out will leave us 2(H2O) on the right only. Then we add the H\'s of the rxns.
 
 Now consider reaction for Eliminating N2O
(N2O) + (3H2) ---> (N2H4) + (H2O) H -317 kJ   multiplying this by 3 and reversed so
 3(N2H4) + 3(H2O) ---> 3(N2O) + 9(H2)     H +951 kJ
Then add with reaction A
 N2H4 + 3(N2O) ---> 4(N2) + 2(H2O) + 1/2(O2) H -867 kJ
 +
 3(N2H4) + 3(H2O) ---> 3(N2O) + 9(H2) H +951 kJ
 --------------------------------------...
 4(N2H4) + H2O ---> 4(N2) + 1/2(O2) + 9(H2) H 84 kJ ---------------(B)
 
 both side 3(N2O) get cancel and two H2O from both side and one is remaining at left side.
 Now consider reaction to Eliminating H2
(H2) + 1/2(O2) ---> (H2O) H -286 multiplying by 9 we get
9(H2) + 9/2(O2) ---> 9(H2O) H -2574 kJ then add it with reaction B
 
 4(N2H4) + H2O ---> 4(N2) + 1/2(O2) + 9(H2) H 84 kJ
 +
 9(H2) + 9/2(O2) ---> 9(H2O) H -2574 kJ
 --------------------------------------...
 4(N2H4) + 4(O2) ---> 4(N2) + 8(H2O) H -2490 kJ ------------(C )
 
 both side 9(H2) get cancel ½ O2 right side. From right side and H2o from left side get cancel. So we get new reaction (C)
 
 now devide reaction C we have get our proposed reaction that is
 (N2H4) + 4(O2) ---> 4(N2) + 8(H2O) H -2490 kJ ------------(C ) devide by 4
 N2H4 + O2 ---> N2 + 2(H2O) H -622.5 kJ

