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nikdorinn [45]
2 years ago
10

When 3.93 grams of lactic acid, CHoOs(s), are burned in a bomb

Chemistry
1 answer:
aliya0001 [1]2 years ago
5 0

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 <u>-716.2 kJ</u>

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 ≈ <u>-716.2 kJ</u>.

Learn more here:

brainly.com/question/13185938

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Initial (M)              0.08            0.04                            0

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Equilibrium (M)      0.04             0                                0.04

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