Standard Heat of Reaction — Oxidation of Toluene to Benzaldehyde: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2019, Q1

Question A1: Standard Heat of Reaction — Oxidation of Toluene to Benzaldehyde (Part A — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Oxidation $\mathrm{C_6H_5CH_3\,(g)+O_2\,(g)\rightarrow C_6H_5CHO\,(g)+H_2O\,(g)}$ (balanced as written: 1 mol O$_2$). Benzaldehyde $M=106.13$ g/mol. Standard heats of formation from Felder App. B (open-book): $\Delta H^\circ_{f,\mathrm{CO_2(g)}}=-94.05$, $\Delta H^\circ_{f,\mathrm{H_2O(l)}}=-68.32$ kcal/mol (consistent with the given $\Delta H^\circ_{f,\mathrm{H_2O(g)}}=-57.8$).

Datum	Value
Gross heat of combustion, liquid benzaldehyde @ 18 °C	−841.3 kcal/gmol
Normal boiling point of benzaldehyde	179 °C
$\Delta H_{vap}$ @ 179 °C	86.48 cal/g
$C_p$ liquid benzaldehyde	0.428 cal/g·°C
$C_p$ benzaldehyde vapour	31 cal/mol·°C
$C_p$ liquid H$_2$O / CO$_2$ / O$_2$	18 / 8.87 / 7.0 cal/mol·°C
$\Delta H^\circ_{f}$ H$_2$O(g) / toluene(g)	−57.8 / +11.95 kcal/mol

Find. the standard heat of reaction $\Delta H^\circ_{rxn}$ (at 25 °C) for the oxidation of toluene to benzaldehyde vapour.

Figure A1 — Thermochemical route to $\Delta H^\circ_{f}$ of benzaldehyde vapour: its gross heat of combustion (Kirchhoff-corrected to 25 °C) inverts to the liquid-phase formation enthalpy, to which the 25 °C heat of vaporisation is added. Hess’s law then closes the target reaction.

Approach. Benzaldehyde is the only species without a tabulated $\Delta H^\circ_f$; recover it from its gross heat of combustion (Kirchhoff-correct 18→25 °C, invert by Hess to the liquid value, then add the 25 °C heat of vaporisation), and finally apply Hess’s law to the target reaction.

Liquid molar heat capacity. Convert the mass-basis value with $M=106.13$ g/mol: $$C_{p,\ell}=0.428\times106.13=45.42\ \text{cal/mol}\cdot\text{}^\circ\text{C}.$$
Correct the combustion heat 18 → 25 °C (Kirchhoff). For $\mathrm{C_7H_6O(l)+8O_2\rightarrow7CO_2+3H_2O(l)}$, $\Delta C_p=[7(8.87)+3(18)]-[45.42+8(7.0)]=14.67$ cal/mol·°C, so $$\Delta H_c(25)= -841.3+\tfrac{14.67(7)}{1000}= -841.2\ \text{kcal/mol}.$$ The correction is only 0.1 kcal — combustion enthalpies are weak functions of $T$ here.
Invert to $\Delta H^\circ_{f}$ of liquid benzaldehyde (Hess). $$\Delta H^\circ_{f,\ell}=7(-94.05)+3(-68.32)-\Delta H_c(25)= -22.11\ \text{kcal/mol}.$$
Heat of vaporisation at 25 °C. With $\Delta H_{vap}(179)=86.48\times106.13/1000=9.18$ kcal/mol and a liquid-heat/vapour-cool path ($\Delta T=154$ °C), $$\Delta H_{vap}(25)=C_{p,\ell}\Delta T+\Delta H_{vap}(179)-C_{p,v}\Delta T=\tfrac{45.42(154)+9178-31(154)}{1000}=11.40\ \text{kcal/mol}.$$
Formation enthalpy of the vapour. $$\Delta H^\circ_{f,\mathrm{C_7H_6O(g)}}= -22.11+11.40 = \boxed{-10.71\ \text{kcal/mol}}.$$
Heat of reaction (Hess’s law). With $\Delta H^\circ_{f}(\mathrm{O_2})=0$, $$\Delta H^\circ_{rxn}=[\Delta H^\circ_{f,\mathrm{C_7H_6O(g)}}+\Delta H^\circ_{f,\mathrm{H_2O(g)}}]-\Delta H^\circ_{f,\mathrm{tol(g)}}=(-10.71-57.8)-11.95=\boxed{-80.5\ \text{kcal/mol}}.$$ Equivalently $-336.7$ kJ/mol — the partial oxidation is strongly exothermic.

Quantity	Result
$\Delta H^\circ_{f}$ benzaldehyde, liquid	−22.11 kcal/mol
$\Delta H_{vap}$ benzaldehyde @ 25 °C	+11.40 kcal/mol
$\Delta H^\circ_{f}$ benzaldehyde, vapour	−10.71 kcal/mol
Standard heat of reaction	−80.5 kcal/mol (−336.7 kJ/mol)

Concept & Theory

Topic. Heat of reaction built from a heat of combustion via Hess’s law.

Key relations. $\Delta H^\circ_{rxn}=\sum\nu_i\Delta H^\circ_{f,i}$; combustion–formation inversion $\Delta H^\circ_{f}=\sum\nu\Delta H^\circ_{f,\text{products}}-\Delta H_c$; Kirchhoff $\Delta H(T_2)=\Delta H(T_1)+\Delta C_p\Delta T$; phase-change path for $\Delta H_{vap}(T)$.

Why this works. Benzaldehyde is the one species with no tabulated formation enthalpy, so its value must be manufactured from the measured (gross) heat of combustion. ‘Gross’ means the product water is liquid, which is why $\Delta H^\circ_{f,\mathrm{H_2O(l)}}$ (not the vapour value) enters the inversion. Because all enthalpies are state functions, the small 18→25 °C and 179→25 °C corrections can be applied along any convenient path.

Common pitfall. Using the vapour water enthalpy in a ‘gross’ combustion (or forgetting to add $\Delta H_{vap}$ to reach the vapour product required by the reaction) throws the answer off by tens of kcal.

Source. Felder, Rousseau & Bullard, Elementary Principles §9.5–9.6 (heats of reaction, formation and combustion; Hess’s law).

Question 1 of 6: Standard Heat of Reaction — Oxidation of Toluene to Benzaldehyde

Part A — Process Mass & Energy Balances

Question A1: Standard Heat of Reaction — Oxidation of Toluene to Benzaldehyde (Part A — equal value)