Question

When 3.93 grams of lactic acid, CHoOs(s), are burned in a bomb
calorimeter that has a total heat capacity of 10.80 kJ-K-1, the
temperature of the calorimeter increases by 5.43 K. Assume AHrxn
=
AErxn. Given that AH°f of H¿O(I) = -285.8 kJ-mol-1 and AH°f of CO:(g)
= -393.5 kJ• mol-1, calculate the AH°f of lactic acid.

Answer 1

The heat released in the combustion of lactic acid is absorbed by the

calorimeter and in the decomposition of the lactic acid.

ΔH°f of lactic acid is approximately -716.2 kJ

Reasons:

Known parameters are;

Mass of the lactic acid = 3.93 grams

Heat  capacity of the bomb calorimeter = 10.80 kJ·K⁻¹

Change in temperature of the calorimeter, ΔT = 5.34 K

ΔHrxn = ΔErxn

ΔH°f of H₂O(l) = -285.8 kJ·mol⁻¹

ΔH°f of CO₂(g) = -393.5 kJ·mol⁻¹

The chemical equation for the reaction is presented as follows;

• C₃H₆O₃ + 2O₂ → 3CO₂ + 3H₂O

The heat of the reaction = 10.80 kJ·K⁻¹ × 5.34 K = 57.672 kJ

Molar mass of C₃H₆O₃ = 90.07 g/mol

Number of moles of C₃H₆O₃ = $\dfrac{3.93 \, g}{90.07 \, g/mol}$ = 0.043633 moles

Number of moles of CO₂ produced = 3 × 0.043633 moles = 0.130899 moles

Heat produced = 0.130899 mole × -285.8 kJ·mol⁻¹ = -37.4109342 kJ

Moles of H₂O produced = 0.130899 moles

Heat produced = 0.130899 mole × -393.5 ≈ -51.51 kJ

Therefore, we have;

Heat absorbed by the lactic acid = ΔH°f of H₂O + ΔH°f of CO₂ + Heat absorbed by the calorimeter

Which gives;

Heat absorbed by lactic acid  = -37.4109342 kJ - 51.51 kJ + 57.672 kJ ≈ -31.249 kJ

The heat absorbed by the lactic acid ≈ -31.249 kJ

• $\Delta H^{\circ}f \ of \ C_3H_6O_3 = \dfrac{-31.249}{0.043633} \approx -716.2$

ΔH°f of C₃H₆O₃ ≈ -716.2 kJ

Heat of formation of lactic acid ≈ -716.2 kJ.

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