Answer
$\frac{1}{2} N_2(g) + O_2(g) --> NO_2(g)$
$\Delta H^{\circ}_f (NO_2(g)) = 33.84 kJ/mol$
$\Delta H^{\circ}_f (N_2(g)) = 0 kJ/mol$
$\Delta H^{\circ}_f (O_2(g)) = 0 kJ/mol$
Work Step by Step
1. Write the unbalanced equation:
$N_2(g) + O_2(g) --> NO_2(g)$
2. Balance it:
$\frac{1}{2}N_2(g) + O_2(g) --> NO_2(g)$
3. Check the book's final pages, and find the $\Delta H^{\circ}_f$:
$\Delta H^{\circ}_f (NO_2(g)) = 33.84 kJ/mol$
$\Delta H^{\circ}_f (N_2(g)) = 0 kJ/mol$
$\Delta H^{\circ}_f (O_2(g)) = 0 kJ/mol$
