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Author Question: H2S(g) + O2(g) H2O (l) + SO2(g) rH = -562.3 kJ/mol 1/8 S8(s) + O2(g) SO2(g) rH = -297.0 kJ/mol ... (Read 21 times)

asmith134

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H2S(g) + O2(g) → H2O (l) + SO2(g)  Δr = -562.3 kJ/mol
1/8 S8(s) + O2(g) → SO2(g)  Δr = -297.0 kJ/mol
H2(g) + O2→ H2O (l)  Δr = -266.9 kJ/mol

Compute Δr for H2(g) + 1/8 S8 (s) → H2S(g) in kJ/mol.
◦ 562.3 kJ/mol
◦ 276.6 kJ/mol
◦ 265.3 kJ/mol
◦ 20.5 kJ/mol
◦ -1.6 kJ/mol


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Marked as best answer by asmith134 on Jul 8, 2021

Dinolord

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asmith134

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Reply 2 on: Jul 8, 2021
Thanks for the timely response, appreciate it


rachel

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Reply 3 on: Yesterday
Great answer, keep it coming :)

 

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