Question 2 of 6: Theoretical (Adiabatic) Flame Temperature

Question 2: Theoretical (Adiabatic) Flame Temperature (Part A — equal value)

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Given. Fuel gas 30 mol% CO, 70 mol% N₂; reaction $\text{CO}+\tfrac12\text{O}_2\rightarrow\text{CO}_2$; 100% excess air; reactants at 298 K. $\Delta H^\circ_{f,\text{CO}_2}=-393.7$, $\Delta H^\circ_{f,\text{CO}}=-110.6$ kJ/mol. Mean $C_p$ (J/mol·K): CO₂ 50.1, O₂ 33.3, N₂ 31.5. Air is 21% O₂ / 79% N₂. Basis: 100 mol fuel gas.

Find. The theoretical (adiabatic) flame temperature $T$ of the product gas.

Approach. For an adiabatic reactor the enthalpy released by reaction at 298 K equals the sensible heat that warms the product gas from 298 K to $T$; take a reaction–then–heat path with $Q=0$.

1. Heat of reaction. Per mole of CO, $\Delta H_r=\Delta H^\circ_{f,\text{CO}_2}-\Delta H^\circ_{f,\text{CO}}=-393.7-(-110.6)=-283.1$ kJ/mol. For 30 mol CO: $$Q_{\text{rxn}} = 30\times283.1 = 8493\ \text{kJ}=8.493\times10^{6}\ \text{J (released at 298 K)}.$$
2. Product inventory. All CO oxidises: CO₂ $=30$ mol; excess O₂ $=30-15=15$ mol; total N₂ $=70+112.9=182.9$ mol. These products absorb the released heat.
3. Total heat capacity of the products. $$\textstyle\sum n_iC_{p,i}=30(50.1)+15(33.3)+182.9(31.5)=1503+499.5+5760=7762.5\ \text{J/K}.$$
4. Adiabatic energy balance. With $Q=0$, released enthalpy warms the products from 298 K to $T$: $$Q_{\text{rxn}}=\Big(\textstyle\sum n_iC_{p,i}\Big)(T-298)\;\Rightarrow\;T-298=\frac{8.493\times10^{6}}{7762.5}=1094\ \text{K},$$ $$\boxed{T_{\text{flame}}\approx 1392\ \text{K}\ (\approx 1119\,{}^{\circ}\text{C}).}$$

The large ballast of inert N₂ (fuel diluent plus the 79% of a doubled air supply) is what keeps the flame temperature moderate; without excess air and fuel nitrogen it would run several hundred kelvin hotter.