Question

Use the thermochemical equations shown below to determine the enthalpy for the reaction:

C2H6O(l)+3O2(g)2CO2(g)+3H2O(l)

2C2H4O(l)+2H2O(l)2C2H6O(l)+O2(g) H= +203.5kJ 2CO2 (g)+2H2O(l)C2H4O(l)+5/2O2(g) H= +583.5kJ

Answer 1

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