Calculate enthalpy change for the change 8S(g) → S8(g), given that H2S2(g) → 2H(g) + 2S(g), Δ H =239.0 k cal mol-1 H2S2(g) → 2H(g) + S(g), Δ H = 175.0 k cal mol-1

Question

Calculate enthalpy change for the change 8S(g) → S8(g), given that H2S2(g) → 2H(g) + 2S(g), Δ H =239.0 k cal mol-1 H2S2(g) → 2H(g) + S(g), Δ 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

Tardigrade Admin · Accepted Answer

Δ H S-S + 2Δ HH-S = 239 2 Δ HH-S = 175 Hence, Δ H S-S = 239-175 = 64 kcal mol-1 Then, Δ H for 8S(g) → S8(g) is 8× (-64) =-512 kcal

Q. Calculate enthalpy change for the change $8 S (g) \to S_{8} (g)$ , given that
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g)$ ,
$Δ H = 239.0 k c a l m o l^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g)$ ,
$Δ H = 175.0 k c a l m o l^{- 1}$

+ 512.0 k cal

- 512.0 k cal

508.0 k cal

-508.0 k cal

$Δ H_{S - S} + 2Δ H_{H - S} = 239$
$2Δ H_{H - S} = 175$
Hence,
$Δ H_{S - S} = 239 - 175 = 64 k c a l m o l^{- 1}$
Then, $Δ H$ for $8 S_{(g)} \to S_{8 (g)}$ is
$8 \times (- 64) = - 512 k c a l$

Q. Calculate enthalpy change for the change 8S(g)→S8​(g), given that H2​S2​(g)→2H(g)+2S(g), ΔH=239.0kcalmol−1 H2​S2​(g)→2H(g)+S(g), ΔH=175.0kcalmol−1