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

  

‒292 kJ/mol

   

292 kJ/mol

   

‒146 kJ/mol

50.6 kJ/mol

‒50.6 kJ/mol

Answer 1

Solution :-

Given data

Standard enthalpy of formation of the water is given Delta H f H2O = -241.8 kJ per mol

Enthalpy change of reaction Delta H rxn = -534.2 kJ

Delta H f N2H4 = ?

Values of the Delta Hf of N2(g) and O2(g) = 0.00 kJ ( they are in elemental form)

Using the Hess's law formula we can calculate the enthalpy change of formation of the N2H4(l)

Delta H rxn = sum of Delta Hf product - sum of Delta Hf reactant

lets make the set up

Delta H rxn = [H2O *2 ] -[ N2H4*1]

lets put the values in the formula

-534.2 kJ = [-241.8 kJ * 2] - [ N2H4]

-534.2 kJ = -483.6 kJ - [N2H4]

[N2H4] = 534.2 kJ - 483.6 kj

[N2H4] = 50.6 kJ /mol

Therefore the standard enthalpy of formation of the N2H4(l) = 50.6 kJ/ mol