Answer to Question #286886 in Chemistry for michelle

Question

When 10.0 g of liquid nitrogen is vapourized, 4000 J of energy is released. What is the molar enthalpy of vapourization for liquid nitrogen? 

1. – 11.2 kJ/mol
2. + 5.6 kJ/mol
3. 0 kJ/mol
4. + 11.2 kJ/mol
5. – 5.6 kJ/mol

Answer 1

Solution:

Calculate the number of moles of liquid nitrogen (N₂):

The molar mass of N₂ is 28.0 g/mol

Hence,

(10.0 g N₂) × (1 mol N₂ / 28.0 g N₂) = 0.357 mol N₂

Calculate the molar enthalpy of vaporization (ΔH_v) for liquid nitrogen:

ΔH_v = energy / mol of substance

ΔH_v(N₂) = (4000 J) / (0.357 mol) = 11204.48 J/mol = 11.2 kJ/mol

The molar enthalpy of vaporization (ΔH_v) for liquid nitrogen (N₂) is 11.2 kJ/mol

Answer: 4. +11.2 kJ/mol