Enthalpy (Chemistry)

Question

Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values:

N₂H₄(l) + H₂(g) → 2NH₃(g)

N₂H₄(l) + CH₄O(l) → CH₂O(g) + N₂(g) + 3H₂ (g)        ΔH = -37 kJ

N₂(g) + 3H₂(g) → 2NH ₃(g)                                               ΔH = -46 kJ

CH₄O(l) → CH₂O(g) + H ₂(g)                                             ΔH = -65 kJ

Answer 1

Another Hess' Law

N2H4(l) + H2(g) ==> 2NH3(g) ... TARGET EQUATION

Given:

eq.1. N₂H₄(l) + CH₄O(l) → CH₂O(g) + N₂(g) + 3H₂ (g)      ΔH = -37 kJ

eq.2. N₂(g) + 3H₂(g) → 2NH ₃(g)          ΔH = -46 kJ

eq.3. CH₄O(l) → CH₂O(g) + H ₂(g)       ΔH = -65 kJ

Procedure:

copy eq.1. N₂H₄(l) + CH₄O(l) → CH₂O(g) + N₂(g) + 3H₂ (g)      ΔH = -37 kJ

reverse eq.3. CH2O(g) + H2(g) ==> CH4O(l) ∆H = +65 kJ

copy eq.2. N₂(g) + 3H₂(g) → 2NH ₃(g)          ΔH = -46 kJ

Add them up to get...

N₂H₄(l) + CH₄O(l) + CH2O(g) + H2(g) + N₂(g) + 3H₂(g) => CH₂O(g) + N₂(g) + 3H₂ (g) + CH4O(l) + 2NH ₃(g)

Combine and cancel like terms to get...

N2H4(l) + H2(g) ==> 2NH3(g) ... TARGET EQUATION

Add up ∆H values to get...

∆H = -37 kJ + 65 kJ + -46 kJ

∆H = -18 kJ