Question

Given the following heats of formation (ΔH°_f, 25°C) and heats of combustion (ΔH°_c, 25°C) data, determine the heat of reaction for the liquid-phase esterification of lactic acid (C₃H₆O₃) with ethanol (C₂H₅OH) to form ethyl lactate (C₅H₁₀O₃) and liquid water at 25°C and 1 atm.

C₃H₆O₃ + C₂H₅OH --> C₅H₁₀O₃ + H₂O (liquid)

The table below shows the standard heat of combustion and heat of reaction.

Compound	ΔH°c (25°C, 1 atm) kJ/mol	ΔH°f (25°C, 1 atm) kJ/mol
C3H6O3	-	-687.0
C2H5OH	-	-277.6
C5H10O3	-2685	-
H2O (liquid)	-	-285.8
CO2 (gas)	-	-393.5
O2 (gas)	-	0

Answer 1

Ans :-  ΔH⁰_r = - 32.7 KJ/mol

Explanation :-

Step 1. Calculation of ΔH⁰_f of C₅H₁₀O₃ :-

The combustion equation of C₅H₁₀O₃ is :

C₅H₁₀O₃ + O₂ ---------------> 5 CO₂ + 5 H₂O , ΔH⁰_c = ΔH⁰_r = - 2685 KJ/mol (Given)

We know

ΔH⁰_r = Σ ΔH⁰_f of Products - Σ ΔH⁰_f of Reactants

- 2685 KJ/mol = [ 5 ΔH⁰_f of CO₂ + 5 ΔH⁰_f of H₂O] - [ ΔH⁰_f of C₅H₁₀O₃ + 6  ΔH⁰_f of O₂]

- 2685 KJ/mol = [ 5 (-393.5 KJ/mol) + 5 (-285.8 KJ/mol)] - [ ΔH⁰_f of C₅H₁₀O₃ + 0 ]

- 2685 KJ/mol = [ -1967.5 KJ - 1429 KJ ] - ΔH⁰_f of C₅H₁₀O₃

ΔH⁰_f of C₅H₁₀O₃ = 2685 KJ - 3396.5 KJ

ΔH⁰_f of C₅H₁₀O₃ = -711.5 KJ

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Step 2. Calculation of ΔH⁰_r :-

Given reaction is

C₃H₆O₃ + C₂H₅OH --> C₅H₁₀O₃ + H₂O (liquid) , ΔH⁰_r = ?

ΔH⁰_r = Σ ΔH⁰_f of Products - Σ ΔH⁰_f of Reactants

ΔH⁰_r = [  ΔH⁰_f of C₅H₁₀O₃ +  ΔH⁰_f of H₂O] - [ ΔH⁰_f of C₃H₆O₃ + ΔH⁰_f of C₂H₅OH ]

ΔH⁰_r = [ -711.5 KJ/mol - 285.8 KJ/mol ] - [ -687.0 KJ/mol -277.6 KJ/mol ]

ΔH⁰_r = - 997.3 KJ/mol + 964.6 KJ/mol

ΔH⁰_r = - 32.7 KJ/mol