Question

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

Answer 1

Step-by-step 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:

`\Delta H_(reac)\text{ = 2}*358+498\text{ = 1214kJ/mol}`

Bond energy of products:

`\Delta H_(prod)\text{ = 2}*799\text{ = 1598kJ/mol}`

Total change in bond energy:

`\Delta H_(reaction)\text{ = }\Delta H_(reac)\text{ - }\Delta H_(prod)\text{ = 1214 - 1598 = -384kJ/mol}`

Answer:

Enthalpy of formation = -384kJ/mol