Question

31. A sample of the sugar D-ribose of mass 0.727 g was placed in a calorimeter and then ignited in the presence of excess oxygen. The temperature rose by 0.910 K.. In a separate experiment in the same calorimeter, the combustion of 0.917 g of benzoic acid, for which the internal energy of combustion is - 3226 kJ mol-1, gave a temperature rise of 1.940 K. Calculate the enthalpy of formation of D-ribose.

33. The mean bond enthalpies of C-C, C-H, C=O, and O-H bonds are 348, 412, 743, and 463 Kj mol-1, respectively. the combustion of a fuel such as octane is exothermic because relativley weak bonds break to form relatively strong bonds, use this information to justify why glucose has a lower specific enthalpy than the lipid decanoic acid even though these compounds have similar molar masses.

If you could answer both questions, please, that would be greatly appreciated.

Answer 1

The calorimeter constant is determined by the benzoic acid data

$\Delta$U = (0.917g/122.12g/mol)x -3226 kJ/mol

$\Delta$U = 24.22 kJ

$\Delta$T = 1.940K

C = 24.22/1.94 = 12.486kJ/K

for D-ribose

$\Delta$U = -C x $\Delta$T = -12.486 kJ/K x 0.910K

$\Delta$_fU = -12.486 x 0.910 x 150.13 g/mol/0.727 g = -2346.5 kJ/mol

Enthapy of formation calculation

C₅H₁₀O₅(s) + 5 O₂(g) $\rightarrow$ 5CO₂(g) + 5 H₂O(l)

                                                                                     $\Delta$H

5CO₂(g) + 5 H₂O(l)$\rightarrow$C₅H₁₀O₅(s) + 5 O₂(g)                2346.5

formation of 5 CO2(g)                                               5 x (-393.5)

formation of 5 H2O(l)                                                5 x (-285.8)

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5C(s) +5H2(g) + 2/5 O2(g)   $\rightarrow$C₅H₁₀O₅(s)         -1050 kJ/mol

enthalpy of formation is -1050 kJ/mol