Calculate the enthalpy (∆H) for the following reaction:
1) H2O (g) + CO (g) = CO2 (g) + H2 (g)
2) C2H2 (g.) + 5 / 2O2 (g.) → 2CO2 (g.) + H2O (l.)
CO(g) = -99 kJ/mol.
1. ∆H= -393.5 -(-99 + (-241.6))= -52,9 kJ/mol.
2. ∆H= 2 x -393.5 + (-241.6) - 226.7 = 1 255,3 kJ/mol.
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