First slide

Enthalpy of Combustion

Question

ΔH for S_α → S_β is + 0.32 KJ/ mole. Then,
ΔH for 1/8S_β + O_2(g) → SO_2(g) is
[ Δ_fH (SO₂) is -296 KJ/mole]

Easy

Solution

$\begin{array}{l} {\Delta _r}{H^o} = {\Delta _f}H(S{O_2}) - [\frac{1}{8}{\Delta _f}{H^o}({S_\beta }) + {\Delta _f}{H^o}({O_2})]\\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \, - 296 - [\frac{1}{8}( + 0.32) + 0]\\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 296 - 0.04 = - 296.04KJ \end{array}$

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