Question 4 of 7: Adiabatic Flame Temperature of Acetylene — Oxygen vs. Air
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).
Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state enthalpy/entropy data used below; supporting 298 K entropies for the elements, CO, CO₂, H₂, H₂O(g) and Cu₂O/Cr₂O₃/TiO₂/SiO₂ from the NIST–JANAF Thermochemical Tables.
Question 4: Adiabatic Flame Temperature of Acetylene — Oxygen vs. Air (20 marks)
Given. $\text{C}_2\text{H}_2+2.5\,\text{O}_2=2\text{CO}_2+\text{H}_2\text{O}$, all at 298 K initially. $\Delta H^\circ_f$: C₂H₂ = +226.7 kJ/mol, CO₂ = −393.5 kJ/mol, H₂O = −241.8 kJ/mol. $C_p(T)$ for CO₂, H₂O, N₂ as given above. Air: 21% O₂, 79% N₂ by mole.
Find. (a) Adiabatic flame temperature $T_f$ burning in pure, stoichiometric O₂ (no N₂). (b) $T_f$ burning stoichiometric O₂ supplied as air (N₂ along for the ride).
Approach. Combustion is adiabatic and at constant pressure, so all of the reaction's exothermic $\Delta H^\circ_{298}$ goes into heating the product gases from 298 K to $T_f$: $-\Delta H^\circ_{rxn,298}=\int_{298}^{T_f}\sum n_iC_{p,i}(\text{products})\,dT$. Part (b) simply adds the diluent N₂ (which absorbs heat but releases none) to the product-heating integral.
Reaction enthalpy at 298 K.
$$\Delta H^\circ_{rxn}=2(-393{,}500)+(-241{,}800)-(226{,}700)=\boxed{-1{,}255{,}500\ \text{J per mol }\text{C}_2\text{H}_2}.$$
(a) Stoichiometric O₂ — products are 2 mol CO₂ + 1 mol H₂O only. Summing their $C_p(T)$: $\sum n_iC_{p,i}=2(18.9+0.079T)+(31.4+0.004T)=69.2+0.162T$. The energy balance
$$-\Delta H^\circ_{rxn}=\int_{298}^{T_f}(69.2+0.162T)\,dT=69.2(T_f-298)+0.081(T_f^2-298^2)$$
is a quadratic in $T_f$; solving $0.081T_f^2+69.2T_f-1{,}283{,}315=0$ gives
$$T_f=\frac{-69.2+\sqrt{69.2^2+4(0.081)(1{,}283{,}315)}}{2(0.081)}=\boxed{3576\ \text{K}\ (3303\,{}^{\circ}\text{C})}.$$
(b) Combustion in air — N₂ dilutes but does not react. The stoichiometric O₂ is 2.5 mol, so it arrives with $n_{N_2}=2.5\times\dfrac{79}{21}=9.405$ mol of inert N₂, which also must be heated from 298 K to $T_f$. Adding its $C_p(T)$ to the product sum:
$$\sum n_iC_{p,i}=69.2+0.162T+9.405(27.9+0.004T)=331.6+0.200T.$$
The same energy-balance integral, $-\Delta H^\circ_{rxn}=\int_{298}^{T_f}(331.6+0.200T)\,dT$, now gives a much lower flame temperature because the N₂ absorbs heat without releasing any:
$$\boxed{T_f=2391\ \text{K}\ (2117\,{}^{\circ}\text{C})}.$$
Check
Both integrals were solved for $T_f$ numerically (the governing equation is a quadratic once $C_p$ is linear in $T$, or root-found directly when a $T^2$ term is present). The large drop from 3576 K (pure O₂) to 2391 K (air) is the standard oxy-fuel-vs-air-fuel result: N₂ is a large parasitic heat sink with no chemical payoff.
