Asample 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.

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

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atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer