1.2 g of sugar (C_5H_10O_5) are placed in a bomb calorimeter at 298 K and ignite
ID: 897813 • Letter: 1
Question
1.2 g of sugar (C_5H_10O_5) are placed in a bomb calorimeter at 298 K and ignited in the presence of oxygen. The temperature increase measured by the calorimeter is Delta T =1.12 K. In a separate experiment in the same calorimeter, 1.5 g of benzoic acid (C_7H_6O_2) are ignited in the presence of oxygen and result in a measured temperature increase DeltaT =2.63 K. The internal energy of combustion of benzoic acid is DeltaU_C= -3251 kJ/mol. Calculate the calorimeter constant. Calculate the internal energy of combustion of sugar C_5H_10O_5. Calculate the enthalpy of combustion of sugar C_5H_10O_5 assuming all gas are perfect. Calculate the enthalpy of formation of sugar C_5H_10O_5 knowing that the enthalpy of formation for CO_2 is DeltaH_1 (CO_2) = -393.51 kJ/mol and that the enthalpy of formation for H_2O is Delta H_f (H_2O) = -285.83 kJ/mol.Explanation / Answer
(i) Calculate Calorimeter constant?
Molar mass of benzoic acid (C7H6O2) = 122 g/mol
1.5g of benzoic acid (C7H6O2) = 1.5/122 = 0.0123 moles
energy of combustion of benzoic acid, DUc(C7H6O2) = -3251 kJ/mol
Heat released by combustion of 0.0123 moles of benzoic acid = 0.0123*3251= 40 kJ
temperature increase, DT = 2.63K
DT*calorimeter constant = 40
calorimeter constant, C = 40/2.63 = 15.21 kJ/K
(ii) dUc of combustion of sugar, (C5H10O5)
Molar mass of sugar(C5H10O5) = 150 g/mol
Mass of sugar(C5H10O5) = 1.2g
Moles of sugar(C5H10O5) = 1.2/150 = 0.008 mol
temperature increase, DT = 1.12K
Total Heat released by combustion, dQ = DT*calorimeter constant = 1.12*15.21=17.0352 kJ
Heat released per mole of sugar = 17.0352/0.008 = 2129.4 kJ/mol
For constant volume, dU = dQ
dUc of combustion of sugar, (C5H10O5) = -2129.4 kJ/mol
(iii) dHc of combustion of sugar, (C5H10O5)
dH = dU + d(P*V) = dU + dng*R*T
At constant volume, dU = dQ
dH = dU + dng*R*T
C5H10O5(s) + 5O2(g) = 5CO2(g) + 5H2O(l)
change in no of moles of reaction per mole of sugar = 5-5=0
dHc = dUc = -2129.4 kJ/mol
(iv) enthalpy of formation of sugar, DHf(C5H10O5),
enthalpy of formation of CO2 is dHf(CO2) = -3.51 kJ/mol. and
enthalpy of formation of H2O is dHf(H2O) = -285.83 kJ/mol
C5H10O5(s) + 5O2(g) = 5CO2(g) + 5H2O(l)
dHc = dHf(products)-dHf(reactants) = 5*dHf(H2O)+5*dHf(CO2) - dHf(C5H10O5)
-2129.4 = 5*(-285.83) +5*(-3.51) - dHf(C5H10O5)
dHf(C5H10O5) = 2129.4 + 5*(-285.83) +5*(-3.51) = 682.7 kJ/mol
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