[mymbb123]

2. Given the enthalpy changes at a certain temperature for reactions (a) and (b), calculate the
enthalpy change for reaction (c). Is reaction (c) endothermic or exothermic at this
temperature?
(a) 3H2(g) + N2(g) = 2NH3(g) H = −92.4 kJ mol−1
(b) 2H2(g) + O2(g) = 2H2O(g) H = −483 kJ mol−1
(c) 4NH3(g) + 3O2(g) =2N2(g) + 6H2O
Reaction (c) is exothermic.
[−2 x(a)] + [3 x (b)] = (c)
4NH3 = 6H2 + 2N2
6H2 + 3O2 =6H2O

4NH3 + 3O2  2N2 + 6H2O (c)
Hc = −2xHa + 3xHb
= −2(−92.4) + 3(−483)
= −1264.2 kJ


I do not get why there is a negaetive sign with the 2. Can anyone explain?

[mjc123]

Well, you've written it out, so why did you do that?
Or did you just copy the book's answer without understanding it?

[Enthalpy]

You can use signs over signs, learn rules for it, and sometimes get the right answer by this mean - it's probably what a professor expects from you, and then do it.

Alternately, you can forget all signs and think. In the reaction to be computed, what is the effect of every heat of formation?
The production of H₂O releases heat: 6*241.8kJ (all gaseous)
The destruction of NH₃ absorbs heat: 4*45.9kJ (all gaseous)
So the reaction releases 6*241.8-4*45.9=1267kJ.
You just have to be clear whether the formation of a particular compound has absorbed or released (Hf<0) heat and account it properly in the reaction to be computed.

This is less systematic than formal operations of signs, possibly less academic hence more difficult to teach, but I make far fewer mistakes that way - especially with heats of formations with so many signs on every side.