Question

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

Answer 1

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:

$\Delta H_{reac}\text{ = 2}\times358+498\text{ = 1214kJ/mol}$

Bond energy of products:

$\Delta H_{prod}\text{ = 2}\times799\text{ = 1598kJ/mol}$

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