Standard enthalpies of fromation are usually computed from the calorimetric data

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 Hf°. 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 Hf°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.

Explanation / Answer

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 Hf°for sucrose use the following expression as follows:

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

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

but

Hf(O2) = 0.

Hf(C12H22O11)   = Ho - 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

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