Given the following reactions and their associated enthalpy changes
CO2 (g) → C (s) + O2 (g) ΔH = 393.5 kJ
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) ΔH = -2044 kJ
H2 (g) + 1/2 O2 (g) → H2O (g) ΔH = -241.8 kJ
calculate the enthalpy change for the following reaction:
4 H2 (g) + 3 C (s)→ C3H8 (g)
Apply Hess Law
invert rxn 1, since C(s) is neede din the left side with 3x coefficeint
3C(s) + 3O2 = 3CO2(g) HRxn = -3*(393.5) = -1180.5
invert rxn 2, since we need C3H8 in the right side
3CO2 + 4H2O --> C3H8 + 5O2 H = 2044
get rid of water via --> multiplucaiton of rxn 3 by 4
4H2 + 2O2 = 4H2O H= 4*(-241.8)
add all
4H2 + 2O2 + 3CO2 + 4H2O + 3C(s) + 3O2 = 3CO2(g) + C3H8 + 5O2 + 4H2O HRxn = -1180.5+ 2044 + 4*(-241.8)
get rid of similar species
4H2 + 3C(s) = C3H8 HRxn = -103.7 kJ
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