Using the table for standard enthalpy of formation, solve 2 CO (g) + O2 (g) --> 2 CO2 (g)




Explanation:
We are given: bond energy of C-O = 358kJ/mol
: bond energy of O2 = 498kJ/mol
: bond energy of CO2 = 799kK/mol
Bond energy of reactants:
[tex]\Delta H_{reac}\text{ = 2}\times358+498\text{ = 1214kJ/mol}[/tex]Bond energy of products:
[tex]\Delta H_{prod}\text{ = 2}\times799\text{ = 1598kJ/mol}[/tex]Total change in bond energy:
[tex]\Delta H_{reaction}\text{ = }\Delta H_{reac}\text{ - }\Delta H_{prod}\text{ = 1214 - 1598 = -384kJ/mol}[/tex]Answer:
Enthalpy of formation = -384kJ/mol