Question

Given the thermochemical equation: 2NaN3(s) → 2 Na(s) + 3N2(g) ∆H˚= +42.7 kJ. What is the value of ∆H˚ or the preparation of 1.50 kg of N2(g)?

Answer 1

According to the given equation, $\Delta{H^0}=+42.7 kJ$ for 3 moles of N₂ that is produced. The mass of 3 moles of N₂ equals $28.01g/mol\times 3mol=84.03g.$

Therefore, for 1.50 kg (or 1500 g) of N₂:

$\Delta{H^0}=\frac{42.7kJ\times1500g}{84.03g}=+762kJ$

Answer: +762 kJ