A sample of the sugar d-ribose (C5H10O5) 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 0.825 g of benzoic acid, for which the internal energy of combustion is - 3251 kJ mol-1, gave a temperature rise of 1.940 K. Calculate the enthalpy of formation of d-ribose.

Question

Sanaa Norman

Chemistry

29 March, 05:17

A sample of the sugar d-ribose (C5H10O5) 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 0.825 g of benzoic acid, for which the internal energy of combustion is - 3251 kJ mol-1, gave a temperature rise of 1.940 K. Calculate the enthalpy of formation of d-ribose.

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Lamar · Answer 1

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