The enthalpy change is calculated from
total bond enthalpies reactants - total bond enthalpies products.
N2(g) + 3H2(g) → 2NH3(g)bonds broken:
1x N≡N 945 kjmol-1
3x H-H 436 kjmol-1bonds formed:
6x N-H 391kjmol-1The enthalpy change of the reaction is
945+3(436)-6(391) = - 93kJmol-1
total bond enthalpies reactants - total bond enthalpies products.
| bond | mean bond enthalpy kJmol-1 | ||
| N≡N | 945 | ||
| H-H | 436 | ||
| N-H | 391 |
1x N≡N 945 kjmol-1
3x H-H 436 kjmol-1bonds formed:
6x N-H 391kjmol-1The enthalpy change of the reaction is
945+3(436)-6(391) = - 93kJmol-1