Calculate enthalpy change for the change 8Sg- S8g- given thatH2S2g- 2Hg - 2Sg - - H - 239-0 k cal mol-1H2S2g- 2Hg - Sg- - H - 175-0 k cal mol-1

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Enthalpy of Solution

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Question

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

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- 512.0 k cal

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508.0k cal

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-508.0 k cal

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Solution

The correct option is B
- 512.0 k cal

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

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

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

The correct option is B - 512.0 k cal ΔHS−S+2ΔHH−S=239 2ΔHH−S=175 Hence, ΔHS−S=239−175=64 kcal mol−1 Then, ΔH for 8S(g)→Sg(g) is 8×(−64)=−512k cal

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

The correct option is B
- 512.0 k cal

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