Sometimes, a reaction can take place in multiple steps instead of just one.
Hess's Law says "the heat released or absorbed in a chemical process is the same whether the
process takes place in one or in several steps."
Example:
A + B => AB
AB + B => AB2
A + 2B => AB2
H1
H2
Htotal =
H1 + H2
Steps to follow:
1. Balance the individual equations (if not balanced already)
2. Flip the equations around if necessary to cancel out terms on opposite sides
3. Changing the equation around requires a sign change of the H for that individual step.
4. One an equation needs to be multiplied by a number, multiple H for that individual
step by that number.
5. Sum up the individual steps.
Example:
C(s, graphite) + O2 → CO2
ΔH = -393.5 kJ/mol
CO2 → C(s, diamond) + O2
ΔH = + 395.41 kJ/mol
C (s, graphite) => C (s, diamond) ΔH = ?
Example:
CH4 + 2O2(g) -> CO2(g) + 2H2O(l)
H2O(l) -> H2O(g)
H = -890 kJ/mol
H = 44 kJ/mol at 298 K
Calculate the enthalpy of the reaction
CH4 + 2O2(g) -> CO2(g) + 2 H2O(g)
H=?
Example:
Given:
C(s) + O2(g) → CO2(g); ΔH = -393.5 kJ/mol
S(s) + O2(g) → SO2(g); ΔH = -296.8 kJ/mol
C(s) + 2 S(s) → CS2(l); ΔH = 87.9 kJ/mol
What is the value for ΔH for the following reaction
CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g)
Example:
N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2 (g)
ΔH = -37 kJ
N2(g) + 3H2(g) → 2NH3(g)
ΔH = -46 kJ
CH4O(l) → CH2O(g) + H2(g)
ΔH = -65 kJ
Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values:
N2H4(l) + H2 => 2NH3(g)