Answer:
The internal energy of combustion of d-ribose = -2127 kJ/mol
The enthalpy of formation of d-ribose = -1269.65 kJ/mol
Explanation:
Step 1: Data given
Mass of d-ribose = 0.727 grams
The temperature rose by 0.910 K
In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is −3251 kJ mol−1, gave a temperature rise of 1.940 K.
Molar mass of benzoic acid = 122.12 g/mol
Step 2: Calculate ΔU for benzoic acid
The calorimeter is a constant-volume instrument so:
ΔU = q
ΔU = (0.825 g/ 122.12 g/mol) * (−3251 kJ /mol)
ΔU = -21.96 kJ
Step 3: Calculate ΔU for d-ribose
c = |q| / ΔT
⇒ with ΔT = 1.940 K
c = 21.96 kJ / 1.940 K
c = 11.32 kJ /K
For d-ribose: ΔU = -cΔT
ΔU = -11.32 kJ/K * 0.910 K
ΔU = - 10.3 kJ
Step 4: Calculate moles of d-ribose
moles ribose = 0.727 grams / 150.13 g/mol
moles ribose = 0.00484 moles
Step 5: Calculate the internal energy of combustion for d-ribose
ΔrU = ΔU / n
ΔrU = -10.3 kJ / 0.004842 moles
ΔrU = -2127 kJ/mol
Step 6: Calculate The enthalpy of formation of d-ribose
The combustion of ribose is:
C5H10O5(s) + 5O2(g) → 5CO2(g) + 5H20(l)
Since there is no change in the number of moles of gas, ΔrH = ΔU
For the combustion of ribose, we consider the following reactions:
5CO2(g) + 5H2O(l) → C5H10O5(s) +5O2(g) ΔH = -2127 kJ/mol
C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ/mol
H2(g) + 1/2 O2(g) → H2O(l) ΔH = -285.83 kJ/mol
ΔH = 2127 kJ/mol + 5(-393.5 kJ/mol) + 5(-285.83 kJ/mol)
ΔH = 2127 kJ/mol - -1967.5 kJ/mol - 1429.15 kJ/mol
ΔH = -1269.65 kJ/mol
The internal energy of combustion of d-ribose = -2127 kJ/mol
The enthalpy of formation of d-ribose = -1269.65 kJ/mol
