Question

The enthalpy of combustion of benzoic acid (C6H5COOH) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be – 3226.7 kJ/mol. (a) When 0.4654 g of a-D-glucose (C6H12O6) was oxidized the temperature rose from 21.22 °C to 22.28 °C. Calculate the enthalpy of combustion of glucose and the value of DUrxn for the combustion. (Hint: Be sure to write a balanced equation for each reaction first and assume the product H2O is a liquid).

Answer 1

equation of the reaction is given by C₆H₁₂O₆ (s) + 6O₂(g) à 6CO₂(g) + 6H₂O(l)

q = C D T = n_glu D _cU ^o gives D _cU ^o = (6.81 kJ/K) 1.06K (180 g/mol)/0.4654 g or D _cU ^o = - 2792 kJ/mol

Note that D _cU ^o + D (pV) = D _cU ^o + D (nRT)^o = D _cH ^o (assuming a perfect gas). Since D n = net change in #moles of gas = 0, D _cU ^o = D _cH ^o.

Note D _cH ^o = - 2808 kJ/mol in A1.1

Find D _fH ^o for 6C(s) + 6H₂(g) + 3O₂(g) à C₆H₁₂O₆ (s)

6CO₂(g) + 6H₂O(l) à C₆H₁₂O₆ (s) + 6O₂(g) D _cH ^o = +2792 kJ/mol

6C(s) + 6O₂(g) à 6CO₂(g) 6 x D _fH ^o = - 2361

6H₂(g) + 3O₂(g) à 6H₂O(l) 6 x D _fH ^o = -1715

Add these 3 eqns together to get the enthalpy of formation for glucose

D _fH ^o = - 1283 kJ/mol Note D _fH ^o = - 1274 kJ/mol in A1.1