Question

What is the enthalpy change for the formation of hydrazine, N2H4(l), from its elements? N2(g) + 2H2(g) → N2H4(l)

Use the following reactions and enthalpy changes:
N2H4(l) + O2(g) → N2(g) + 2H2O(l) H = −622.2 kJ
H2(g) + 1 2 O2(g) → H2O(l) ⃤H = −285.8 kJ

Answer 1

The enthalpy change for the formation of liquid hydrazine (N2H4) from its elements is 50.6 kJ/mol, derived from the provided combustion reaction of hydrazine and the formation of water.

To determine the enthalpy change for the formation of hydrazine (N2H4) from its elements, we use the provided reactions and apply Hess's Law. First, we have the combustion of hydrazine:

N2H4(l) + O2(g) → N2(g) + 2H2O(l) ΔH = −622.2 kJ/mol

Secondly, the formation of water from hydrogen and oxygen:

H2(g) + ½ O2(g) → H2O(l) ΔH = −285.8 kJ/mol

Since the reaction for forming water is given per mole of H2O and we have 2 moles in the combustion reaction, we multiply this enthalpy change by 2: 2 x (−285.8 kJ/mol) = −571.6 kJ/mol

We then subtract this value from the enthalpy change given for the combustion of hydrazine: (−622.2) − (−571.6) = −50.6 kJ/mol

The enthalpy change for the formation of liquid hydrazine from its elements (nitrogen and hydrogen gases) is thus 50.6 kJ/mol.

Answer 2

N2H4(l) + O2(g) → N2(g) + 2H2O(l) ∆H = −622.2 kJ/mol

H2(g) + 1/2O2(g) → H2O(l) ∆H = −285.8 kJ/mol

N2(g) + 2H2(g) → N2H4(l) ∆H = +622.2 kJ/mol

2H2(g) + O2(g) → 2H2O(l) ∆H = −2(285.8 kJ/mol) = −571.6 kJ/mol

∆H° = [∆H°f(N2H4(l))] - [∆H°f(N2(g))] - 2[∆H°f(H2O(l))]

∆H° = [622.2 kJ/mol] - [0 kJ/mol] - 2[-285.8 kJ/mol]

∆H° = 1193.8 kJ/mol

The enthalpy change for the formation of hydrazine from its elements is 1193.8 kJ/mol.

(it might be wrong, so sorry)