Calculating Standard Enthalpy Changes

'Calculate the standard enthalpy change for the oxidation of ammonia according to the equation: 4NH3(g) + 5O2(g) ----> 4NO(g) +6H2O(g)
Standard enthalpies of formation are as follows:
4NH3(g)=-49kJmol-1 NO(g)=+90kJmol-1 H2O(g)=-242kJmol-1'

Can someone help me with this? And can they explain it step by step, because I'm really stuck
Thankyou :smile:

Reply 1

theroyalwe

Okay so products in:
4NH3 = -49kJmol-1
5O2 = 0
Total = -49kJmol-1

Products out
4NO = 4 x +90 = 360kJmol-1 (you have 4 moles of NO, so you times it by 4)
6H2O = 6 x -242 = -1452kJmol-1
Total = -1092kJmol-1

Enthalpy change = energy in - energy out = -49 - (-1092) = +1043

anyways i hope thats correct!!!