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?

Question

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

Hello, Haily,

Let's first determine the number of moles contained in 10 grams of N₂. The molar mass of N₂ is 28.0 grams/mole.

(10.0 g)/(28.0 grams/mole) = 0.357 moles N₂.

4000J of energy leaves the system. We can write this as -4.00 kJ,

The molar enthalpy is thus (-4.00kJ)/(0.357 moles N₂) or -11.2 kJ/mole