Question

We burn 7.20g of magnesium in excess nitrogen at constant pressure to form solid magnesium nitride then we bring the reaction mixture to 25 degrees celsius in this process 68.35kj of heat is given off, what is the standard molar enthalpy to form magnesium nitride?

Answer 1

$$3Mg + N_2 \rightarrow Mg_3N_2$$

$n=\frac{m}{M}$

$n(Mg)= \frac{7.209}{24.31}=0.2965 mol$

According to equation $n(Mg):n(Mg_3N_2)=3:1$ , then $n(Mg_3N_2)= \frac{n(Mg)}{3}=\frac{0.2965mol}{3}=0.09883mol$

$$\Delta H_{rxn}=-Q_{rxn}$$

$$\Delta H_f^0= \frac{\Delta H_{rxn}}{n(Mg_3N_2)}= \frac{-68.35 kJ}{0.09883mol}=-691.6 \frac{kJ}{mol}$$