Using Hess's Law, the enthalpy of reaction for the following equation:
2C2H4O(l)+2H2O(l)→22C2H4O(l) + 2H2O(l) → 2C2H6O(l) + O2(g)
C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -685.5 kJ
C2H4O(l) + 5/2O2(g) → 2CO2(g) + 2H2O(l) ΔH = -583.5 kJ
6 years ago
We have to multiply the second equation by 2:
$2C_2H_4O+5O_2->4CO_2+4H_2O$2C2H4O+5O2−>4CO2+4H2O
The enthalpy change for this reaction is multiplied by 2:
ΔH=-1167 kJ
We have to invert the first reaction and multiply it by 2:
$4CO_2+6H_2O->2C_2H_6O+6O_2$4CO2+6H2O−>2C2H6O+6O2
The enthalpy change will then be multplied by -2:
ΔH=1371 kJ
Adding those two values of enthalpy change, we have the enthalpy change of the global reaction:
$2C_2H_4O+2H_2O->2C_2H_6O+O_2$2C2H4O+2H2O−>2C2H6O+O2
Enthalpy change=-1167+1371=204 kJ
6 years ago
Answered By Leonardo F
We have to multiply the second equation by 2:
$2C_2H_4O+5O_2->4CO_2+4H_2O$2C2H4O+5O2−>4CO2+4H2O
The enthalpy change for this reaction is multiplied by 2:
ΔH=-1167 kJ
We have to invert the first reaction and multiply it by 2:
$4CO_2+6H_2O->2C_2H_6O+6O_2$4CO2+6H2O−>2C2H6O+6O2
The enthalpy change will then be multplied by -2:
ΔH=1371 kJ
Adding those two values of enthalpy change, we have the enthalpy change of the global reaction:
$2C_2H_4O+2H_2O->2C_2H_6O+O_2$2C2H4O+2H2O−>2C2H6O+O2
Enthalpy change=-1167+1371=204 kJ