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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]

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– 295.96 KJ

+ 295.96 KJ

– 296.04KJ

– 296.32 KJ

answer is C.

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Detailed 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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