S₈ + 8O₂$\stackrel{ }{\to }$8SO₂, $∆$H = – QKJ. Then $∆$H for SO₂ $\stackrel{ }{\to }$ S + O₂ is

When a thermochemical equation is reversed sign of $∆\mathrm{H}$ becomes opposite.

$∆\mathrm{H}$ for SO₂ $\stackrel{ }{\to }$ S + O₂ is +Q

deviding the equation by 8 on both sides $\frac{∆\mathrm{H}}{8}=\frac{\mathrm{Q}}{8}$