Question

Calculate enthalpy change for the change $8S(g) \to S_8(g)$, given that
$H_2S_2(g) \to 2H(g) + 2S(g)$,
$\Delta H =239.0 \,k \,cal \,mol^{-1}$
$H_2S_2(g) \to 2H(g) + S(g)$,
$\Delta H = 175.0 \,k \,cal \,mol^{-1}$

• A. + 512.0 k cal
• B. - 512.0 k cal
• C. 508.0 k cal
• D. -508.0 k cal

Answer 1

Answer: (B)

$\Delta H _{S-S} + 2\Delta H_{H-S} = 239$
$2 \Delta H_{H-S} = 175$
Hence,
$\Delta H _{S-S} = 239-175 = 64 \,kcal\, mol^{-1}$
Then, $\Delta H$ for $8S_{\left(g\right)} \to S_{8\left(g\right)}$ is
$8\times \left(-64\right) =-512\, kcal$