Question

Use the standard heats of formation from the textbook to calculate delta H for the following unbalanced reaction: C6H12O6 + O2 (g) --> CO2 (g) + H2O (g)

Answer 1

The balanced reaction goes as

C6H12O6 + 6O2 = 6CO2 + 6H2O

The delta H of the reaction can be calculated by subtracting the enthalpy of formation of reactants from that of products .

Enthalpy of formation of :

C6H12O6 = -1271 KJ/mol

O2 = 0 KJ/mol

CO2 = -393.5 KJ/mol

H2O (g) = -241.8 KJ/mol

So now the enthalpy of formation of products :

6x (-393.5) + 6 x (-241.8) = -3811.8 KJ

Enthalpy of formation of reactants :

(-1271) + 6 x 0 = -1271 KJ

Delta H = -3811.8 - (-1271)

= -2540.8 KJ