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Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol   Determine the molar enthalpy (in kJ/mol) for the reaction 6 C (s) + 3 H₂ (g) → C₆H₆ (l).

Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol   Determine the molar enthalpy (in kJ/mol) for the reaction 6 C (s) + 3 H₂ (g) → C₆H₆ (l).

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Using the equations

2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol

2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol

C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol

 

Determine the molar enthalpy (in kJ/mol) for the reaction

6 C (s) + 3 H₂ (g) → C₆H₆ (l).

6C(s) 2 (вон, осе 3 6C(s) + знак т Lo (₁5) ht т + + знаед 1200₂ (9) оло озда в сельд Я H со на 15/0₂ (9) анда соз ед -> 2 (2₂ (9) (са, (2) T 1 11 ась нося ДН Соносех ДН= 6071 - 3135.5 АН - -393.5 =-2341 ДН= 3135. St 2 = +2361 774,5 кЈ
Transcribed Image Text:6C(s) 2 (вон, осе 3 6C(s) + знак т Lo (₁5) ht т + + знаед 1200₂ (9) оло озда в сельд Я H со на 15/0₂ (9) анда соз ед -> 2 (2₂ (9) (са, (2) T 1 11 ась нося ДН Соносех ДН= 6071 - 3135.5 АН - -393.5 =-2341 ДН= 3135. St 2 = +2361 774,5 кЈ
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