Question

3. Use the enthalpies of formation in the table below to answer the following questions.

Substance Enthalpy of Formation (kJ/mol), 298 K

Oxygen (O2)(g) 0

Methane (CH4)(g) -74.8

Carbon Dioxide (CO2)(g) -393.5

Water (H2O)(g) -241.8

Water (H2O)(l) -285.8

a) Calculate the change in enthalpy for the combustion of methane using the values in the table above (assuming that the system remains at 298 K) for the combustion of methane to form carbon dioxide and gaseous water.

b) Repeat this calculation assuming that the product is liquid water.

c) Explain what this difference represents.

Answer 1

The reaction is CH4+2O2---->CO2+2H2O. Two cases need to be considered

Enthalpy change= sum of enhalpy of products- sums of enthalpy of reactants

First case is formation of gaseous water

Enthalpy change= 2*(-241.8)+(-393.5)-(-74.8)=-802.3 Kj

When liquid water is formed

Enthalpy change= 2*(-285.8)+(-393.5)-(-74.8)=-890.3Kj

During change of phase of water some heat is required and this is noting both latent heat of vaporization for fomration of 2 moles of water= (890.3-802.3)/2=44Kj/mol