Question

Standard enthalpies of fromation are usually computed from the calorimetric data referring to reactions, which are easy to carry out in the laboratory. For many compounds, the enthalpy of the reaction with O2 (g) (combustion) is commonly used, together with the standard enthalpies of formation of the combustion products, to compute ΔHf°. For example, sucrose C12H22O11, common table sugar, is found to have a standard enthalpy of combustion of -5640.9 kJ/mole. The standard enthalpies of formation of CO2 (g) and H2O (l) are, respectively, -393.51 and -285.83 kJ/mole. Compute the ΔHf°for sucrose from the standard enthalpies of formation for CO2 (g) and H2O (l). Hint: Write the balanced equations representing the formation of each of the above compounds used in the formation of sucrose.

Answer 1

The reaction is:

C12H22O11(s) + 12O2(g) --- > 12CO2(g) + 11H2O(g)

Given that;

The heats of formation are:
∆Hf(H2O) = -285.83 kJ/mole
∆Hf(CO2) = -393.51 kJ/mole
∆Hf(O2) = 0 kJ/mole (For any element)

To calculate the ΔHf°for sucrose use the following expression as follows:

∆Ho = Σ∆Hf(products) - Σ∆Hf(reactants), then

∆Hf(C12H22O11) +∆Hf(O2)   = ∆Ho - Σ∆Hf(products),

but

∆Hf(O2) = 0.

∆Hf(C12H22O11)   = ∆Ho - Σ∆Hf(products)

Now put all values as follows:
∆Hf(C12H22O11) = -5640.9 kJ/mol - [12 * (-393.51 kJ/mole) + 11 * (-285.83 kJ/mole)]

∆Hf(C12H22O11) = -5640.9 kJ/mol +4722.12 kJ/mol + 3144.13 kJ/mol

∆Hf(C12H22O11) =2225.35 kJ/mol