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 enthalpy (in kJ/mol) for the reaction
6 C (s) + 3 H₂ (g) → C₆H₆ (l).
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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...
2 H₂ (g...
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