The standard molar enthalpy of formation for gaseous H2O is −241.8 kJ/mol. What is the standard molar enthalpy of formation for liquid hydrazine (N2H4)?

Question

The standard molar enthalpy of formation for gaseous H₂O is −241.8 kJ/mol. What is the standard molar enthalpy of formation for liquid hydrazine (N₂H₄)?     

N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g)     ΔH° = ‒534.2 kJ

a) 292 kJ/mol

b) 50.6 kJ/mol

c) -146 kJ/mol

d) -292 kJ/mol

e) -50.6 kJ/mol

Please show steps. I want to understand.

Answer 1

N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g)     ΔH° = ‒534.2 kJ

The ∆Hrx = ∑∆H products - ∑∆H reactants. So let us set up this equation.

∑∆H products = 0 + 2x-241.8 = -483.6 kJ

∑∆H reactants = x + 0 = x

∆Hrx = -534.2 kJ

∆Hrx = ∑∆H products - ∑∆H reactants

-534.2 kJ = -483.6 kJ - x

-x = -50.6 kJ

x = ∆H N₂H₄(l) = 50.6 kJ/mole