Ở điều kiện chuẩn, phản ứng của 2 mol Na (thể rắn) với 1/2 mol O2 (thể khí)

$2{\text{Na}}_{\left(s\right)}+\frac{1}{2}{\text{O}}_{2\left(g\right)}\to {\text{Na}}_2{\text{O}}_{\left(s\right)},{\Delta }_r{H}_{298}^{°}=417.98{\text{\hspace{0.33em}kJ·mol}}^{-1}$

$2{\text{Na}}_{\left(s\right)}+\frac{1}{2}{\text{O}}_{2\left(g\right)}\to {\text{Na}}_2{\text{O}}_{\left(s\right)},{\Delta }_r{H}_{298}^{°}=-417.98{\text{\hspace{0.33em}kJ·mol}}^{-1}$

$4{\text{Na}}_{\left(s\right)}+{\text{O}}_{2\left(g\right)}\to 2{\text{Na}}_2{\text{O}}_{\left(s\right)},{\Delta }_r{H}_{298}^{°}=417.98{\text{\hspace{0.33em}kJ·mol}}^{-1}$

$4{\text{Na}}_{\left(s\right)}+{\text{O}}_{2\left(g\right)}\to 2{\text{Na}}_2{\text{O}}_{\left(s\right)},{\Delta }_r{H}_{298}^{°}=-417.98{\text{\hspace{0.33em}kJ·mol}}^{-1}$