$\Delta H^{\theta }=\sum n\Delta H{}_{tw.}^{\theta }\text{prod.}-\sum n\Delta H{}_{tw.}^{\theta }\text{substr.}=$  

$=2\cdot \Delta H{}_{tw.}^{\theta }{}_{NaN{O}_{2\left(s\right)}}-2\cdot \Delta H{}_{tw.}^{\theta }{}_{NaN{O}_{3\left(s\right)}}=$  

$=2\cdot \left(-359\frac{kJ}{mol}\right)-2\cdot \left(-468\frac{kJ}{mol}\right)=$  

$=-718\frac{kJ}{mol}-\left(-936\frac{kJ}{mol}\right)=$  

$=-718\frac{kJ}{mol}+936\frac{kJ}{mol}=+218\frac{kJ}{mol}$     

Zmiana entalpii tej reakcji to ΔH^θ = +218 kJ/mol.