Standard Heat of Reaction for Toluene Oxidation: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2016, Q2

Question 2: Standard Heat of Reaction for Toluene Oxidation (Part A — 25 marks)

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. The oxidation is C₆H₅CH₃(g) + O₂(g) → C₆H₅CHO(g) + H₂O(g). The only species without a tabulated $\Delta H_f^\circ$ is benzaldehyde ($M=106.12$), so its heat of formation must be built from its measured heat of combustion, C₆H₅CHO(l) + 8O₂ → 7CO₂ + 3H₂O(l).

Datum	Value
$\Delta H_c$ benzaldehyde(l), 18 °C (gross)	−841.3 kcal/mol
$\lambda$ benzaldehyde at 179 °C	86.48 cal/g = 9.18 kcal/mol
$C_p$ benzaldehyde l / v	0.428 cal/g°C (=45.4) / 31 cal/mol°C
$\Delta H_f^\circ$ CO₂ / H₂O(l) (standard tables) / H₂O(g) (given)	−94.05 / −68.32 / −57.8 kcal/mol
$\Delta H_f^\circ$ toluene(g)	+11.95 kcal/mol

Find. The standard heat of reaction $\Delta H_{rxn}^\circ$ at 25 °C (298 K) for the gas-phase oxidation.

Approach. Correct the benzaldehyde combustion to 25 °C (Kirchhoff), invert it to get $\Delta H_f^\circ$ of liquid benzaldehyde, vaporise to the gas, then apply Hess’s law $\Delta H_{rxn}^\circ=\sum\Delta H_{f,prod}^\circ-\sum\Delta H_{f,react}^\circ$.

Combustion to 25 °C. With $\Delta C_p=7C_{p,CO_2}+3C_{p,H_2O(l)}-C_{p,bz(l)}-8C_{p,O_2}=0.01467$ kcal/mol°C, the 7 °C shift is negligible: $$\Delta H_c(25)=-841.3+0.01467(25-18)=-841.2\ \text{kcal/mol.}$$
Heat of formation of liquid benzaldehyde. Reversing the combustion (products − reactants, O₂ contributes nothing): $$\Delta H_{f}^{bz(l)}=7(-94.05)+3(-68.32)-(-841.2)=\boxed{-22.1\ \text{kcal/mol.}}$$
Vaporise to benzaldehyde gas at 25 °C. Bring the 179 °C latent heat back to 25 °C through the liquid/vapour $C_p$ difference: $$\lambda_{25}=9.18+(C_{p,l}-C_{p,v})(179-25)=9.18+(0.0454-0.031)(154)=11.40\ \text{kcal/mol,}$$ $$\Delta H_{f}^{bz(g)}=-22.1+11.40=-10.7\ \text{kcal/mol.}$$
Hess’s law for the oxidation. Now every species has a gas-phase $\Delta H_f^\circ$: $$\Delta H_{rxn}^\circ=\big[\Delta H_f^{bz(g)}+\Delta H_f^{H_2O(g)}\big]-\big[\Delta H_f^{tol(g)}+0\big]=(-10.7-57.8)-11.95=\boxed{-80.5\ \text{kcal/mol.}}$$

The reaction is strongly exothermic (−80.5 kcal/mol ≈ −337 kJ/mol), as expected for a partial oxidation that also forms water vapour. The whole calculation hinges on obtaining benzaldehyde’s formation enthalpy from its combustion data, since it is the one species not in the standard tables.

Quantity	Value
$\Delta H_c$ benzaldehyde(l) at 25 °C	−841.2 kcal/mol
$\Delta H_f^\circ$ benzaldehyde(l)	−22.1 kcal/mol
$\lambda$ benzaldehyde at 25 °C	11.4 kcal/mol
$\Delta H_f^\circ$ benzaldehyde(g)	−10.7 kcal/mol
Standard heat of reaction	−80.5 kcal/mol (−337 kJ/mol)

Concept & Theory

Topic. Hess’s law with heat-of-combustion data, Kirchhoff temperature correction, and phase-change (vaporisation) bookkeeping.

Key relations. Hess’s law $\Delta H_{rxn}^\circ=\sum\nu\,\Delta H_f^\circ$; combustion inverse $\Delta H_f^{fuel}=\sum\Delta H_{f,products}^\circ-\Delta H_c$; Kirchhoff $\Delta H(T_2)=\Delta H(T_1)+\Delta C_p(T_2-T_1)$; latent-heat correction $\lambda_{T_2}=\lambda_{T_1}+(C_{p,l}-C_{p,v})(T_1-T_2)$.

Why this works. Enthalpy is a state function, so any thermochemical path connecting reactants and products gives the same $\Delta H$. Combustion data provide a reliable route to formation enthalpies of compounds (like benzaldehyde) not directly tabulated. Because the reaction is written for gas-phase species, the liquid benzaldehyde must be vaporised, and both the combustion and the latent heat must be referred to the same 25 °C reference state.

Common pitfall. Mixing a “gross” (higher) heating value — which condenses the product water to liquid — with a reaction written for water vapour; the $\Delta H_f^\circ$ used for water must match the phase in each step. Also, sign errors when inverting the combustion reaction.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics, 8th ed., §4.3–4.6 (standard heats of reaction, formation, combustion; temperature dependence).

Question 2 of 6: Standard Heat of Reaction for Toluene Oxidation

Question 2: Standard Heat of Reaction for Toluene Oxidation (Part A — 25 marks)