Question #65789

The combustion of sugar, in a bomb calorimeter is similar to the oxidation of sugar in the body. A student ate three sugar cubes, with the masses of 6.84g, 6.75g and 6.79g.

A) calculate the overall molar enthalpy of oxidation of sugar in the body.

B) using these three sugar cubes as representative of regular-sized cubes, determine the amount of energy released by an average-sized cube.

Expert's answer

Answer on the question #65789, Chemistry / Other

Question:

The combustion of sugar, in a bomb calorimeter is similar to the oxidation of sugar in the body. A student ate three sugar cubes, with the masses of 6.84g, 6.75g and 6.79g.

A) calculate the overall molar enthalpy of oxidation of sugar in the body.

B) using these three sugar cubes as representative of regular-sized cubes, determine the amount of energy released by an average-sized cube.

Solution:

A) The reaction of combustion of sugar is:


C12H22O11+12O212CO2+11H2OC_{12}H_{22}O_{11} + 12O_2 \rightarrow 12CO_2 + 11H_2O


According to Hess's law, molar enthalpy of combustion of sugar is:


ΔHc0(C12H22O11)=12ΔHf0(CO2)+11ΔHf0(H2O)ΔHf0(C12H22O11)\Delta H_c^0(C_{12}H_{22}O_{11}) = 12\Delta H_f^0(CO_2) + 11\Delta H_f^0(H_2O) - \Delta H_f^0(C_{12}H_{22}O_{11})


Enthalpy of formation for sugar, carbon dioxide and water are -2221.2 kJ/mol, -393.51 kJ/mol and -285.830 kJ/mol, respectively. Then, molar enthalpy of oxidation of sugar in the body is:


ΔHc0(C12H22O11)=12(393.51)+11(285.830)+2221.2=5645.05 kJ/mol\Delta H_c^0(C_{12}H_{22}O_{11}) = 12 \cdot (-393.51) + 11 \cdot (-285.830) + 2221.2 = -5645.05\ \text{kJ/mol}


B) Average mass of the sugar cube is:


m=6.84+6.75+6.793=6.79 (g)m = \frac{6.84 + 6.75 + 6.79}{3} = 6.79\ (g)


Molar mass of sugar is 342.2965 g/mol. Then average sugar cube contains the following number of the moles of sugar:


n=mM=6.79 (g)342.2965 (g mol1)=0.01985 moln = \frac{m}{M} = \frac{6.79\ (g)}{342.2965\ (g\ mol^{-1})} = 0.01985\ \text{mol}


So, the amount of energy, released by an average-sized cube is:


Q=nΔHc0(C12H22O11)=0.01985 (mol)5645.05 (kJ/mol1)=112 kJ, or 26.8 kcal.Q = n \cdot \Delta H_c^0(C_{12}H_{22}O_{11}) = 0.01985\ (\text{mol}) \cdot 5645.05\ (\text{kJ/mol}^{-1}) = 112\ \text{kJ, or } 26.8\ \text{kcal}.


Answer: A) 5645.05 kJ/mol-5645.05\ \text{kJ/mol}; B) 112 kJ, or 26.8 kcal112\ \text{kJ, or } 26.8\ \text{kcal}

Thermochemistry data was taken from http://webbook.nist.gov/chemistry/ website.

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