Find the ΔH for the reaction below N2H4(l)+H2(g)→2NH3(g)N2H4(l) +CH4O(l)→CH2O(g) +N2(g) +3H2 (g) ΔH = -37 kJN2(g) + 3H2(g) → 2NH 3(g) ΔH = -46 kJCH4O(l) → CH2O(g) + H 2(g) ΔH = -65 kJ

Answer 1

HESS' LAW

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

Given:

(1) N2H4(l) + CH4O(l) ==> CH2O(g) + N2(g) + 3H2(g) ... ∆H = -37 kJ

(2) N2(g) + 3H2(g) ==> 2NH3(g) ... ∆H = -46 KJ

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

Procedure:

Copy (1): N2H4(l) + CH4O(l) ==> CH2O(g) + N2(g) + 3H2(g) ... ∆H = -37 kJ

Copy (2): N2(g) + 3H2(g) ==> 2NH3(g) ... ∆H = -46 KJ

Reverse (3): CH2O(g) + H2(g) => CH4O(l) ... ∆H = +65 kJ (note to reverse sign)

Add them up and combine and/or cancel like terms to get...

N2H4(l) + CH4O(l) + N2(g) + 3H2(g) + CH2O(g) + H2(g) => CH2O(g) + N2(g) + 3H2(g) + 2NH3(g) + CH4O(l)

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

∆H = -37 kJ + -46 kJ + 65 kJ

∆H = -18 kJ