Use the enthalpies of formation in the table below to answer the following quest
ID: 951142 • Letter: U
Question
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.
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
Explanation / Answer
CH4 (g)+ 2 O2 (g) -----> CO2 (g) + 2 H2O(g)
change in enthalpy = enthalpy of products- enthalpy of reactants
= [-393.5 + 2(-241.8)]-[-74.8 + 0]
= - 802.3 KJ
When H2O is liquid then the enthalpy change of reaction.
CH4 (g)+ 2 O2 (g) -----> CO2 (g) + 2 H2O(l)
change in enthalpy = enthalpy of products- enthalpy of reactants
= [-393.5 + 2(-285.8)]-[-74.8 + 0]
= - 890.3 KJ
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