Exercises — Adiabatic flame temperature — calculation with enthalpies of formation

(b) total enthalpy (and also (d), N₂ mass — it's conserved because N₂ doesn't react, but that's a mass statement, not the thermodynamic key). WHY: Insulated ⇒ $Q=0$ (adiabatic). No moving parts ⇒ no work. Constant pressure ⇒ the First Law reduces to $\Delta H=Q=0$, so $H_{\text{prod}}(T_{\text{ad}})=H_{\text{react}}(298)$. (a) is false — the whole point is that temperature rises. (c) $\Delta H_f^{\circ }$ is a fixed table value per molecule but the mixture's chemical enthalpy changes as bonds rearrange.

${\mathrm{O}}_2=0$, ${\mathrm{N}}_2=0$ (both are pure elements in their standard state — nothing to build), ${\mathrm{H}}_2\mathrm{O}(\mathrm{g})=-241.8$ kJ/mol (negative = energy released when built, because O–H bonds are stable).

Use ${\mathrm{H}}_2\mathrm{O}(\mathrm{g})$ ($-241.8$). At flame temperatures water is steam. Using the liquid value ($-285.8$) secretly credits you the latent heat of condensation ($\approx 44$ kJ/mol) that never happens inside the flame, overstating the released heat.

$\Delta H_{\text{rxn}}^{\circ }=[\Delta H_f^{\circ }(\mathrm{C}{\mathrm{O}}_2)+2\Delta H_f^{\circ }({\mathrm{H}}_2\mathrm{O},\mathrm{g})]-[\Delta H_f^{\circ }(\mathrm{C}{\mathrm{H}}_4)+2\cdot 0]$ $=[-393.5+2(-241.8)]-[-74.8]=-877.1+74.8=-802.3\mathbf{kJ}$. Negative ⇒ exothermic ⇒ heat is available to warm the products. This is just Hess's Law: products-minus-reactants of the formation enthalpies.

$\Delta H_{\text{rxn}}^{\circ }=-241.8-0-0=-241.8$ kJ. One product mole, ${\bar{C}}_p=0.045$. $0.045(T_{\text{ad}}-298)=241.8\Rightarrow T_{\text{ad}}-298=5373\Rightarrow T_{\text{ad}}\approx 5671\mathbf{K}.$ This is unphysically high — a signal that dissociation and rising $C_p$ (ignored here) matter at these temperatures. See Chemical equilibrium and dissociation at high temperature.

Products' total heat capacity: $\sum n_j{\bar{C}}_{p,j}=1(0.055)+2(0.045)+7.52(0.033)=0.055+0.090+0.24816=0.39316\text{kJ/K}.$ $T_{\text{ad}}=298+\frac{802.3}{0.39316}\approx 298+2040.9=2339\mathbf{K}.$ The $7.52{\mathrm{N}}_2$ is the biggest chunk of the denominator — the "silent sponge."

Forecast → lower. Extra cold air adds product moles (1 leftover ${\mathrm{O}}_2$ + extra ${\mathrm{N}}_2$) that soak heat, but no extra fuel means the same numerator $-\Delta H_{\text{rxn}}^{\circ }=802.3$ kJ. Bigger denominator ⇒ lower $T_{\text{ad}}$. N₂: $3\times 3.76=11.28$ mol. Numerator unchanged (leftover ${\mathrm{O}}_2$ has $\Delta H_f^{\circ }=0$, doesn't change $\Delta H_{\text{rxn}}^{\circ }$). $\sum n_j{\bar{C}}_{p,j}=1(0.055)+2(0.045)+1(0.033)+11.28(0.033)=0.055+0.090+0.033+0.37224=\mathrm{0.55024.}$ $T_{\text{ad}}=298+\frac{802.3}{0.55024}\approx 298+1458.1=1756\mathbf{K}.$ Confirmed lower than the stoichiometric $2339$ K. See Stoichiometric vs lean vs rich mixtures.

$\sum n_j{\bar{C}}_{p,j}=1(0.055)+2(0.045)=0.145\text{kJ/K}.$ $T_{\text{ad}}=298+\frac{802.3}{0.145}\approx 298+5533.1=5831\mathbf{K}.$ Removing N₂ raised $T_{\text{ad}}$ from $\approx 2339$ K to $\approx 5831$ K — a jump of $\approx 3492$ K. The nitrogen was absorbing thousands of kelvin of heating capacity. (Real flames cap near 3000 K because dissociation kicks in — the constant-$C_p$ model can't see that.)

Wrong reaction enthalpy: $\Delta H_{\text{rxn}}^{\circ }=[-393.5+2(-285.8)]-[-74.8]=-965.1+74.8=-890.3$ kJ. $T_{\text{ad}}^{\text{wrong}}=298+\frac{890.3}{0.39316}\approx 298+2264.6=2563\mathbf{K}.$ Inflation $=2563-2339=224\mathbf{K}$ too hot — purely from crediting condensation heat that never occurs. Small-looking data error, large temperature error.

N₂: $1.5\times 3.76=5.64$ mol. $\Delta H_{\text{rxn}}^{\circ }=[-110.5+2(-241.8)]-[-74.8]=-594.1+74.8=-519.3\text{kJ}.$ Much less energy released than $-802.3$ (CO still holds combustible energy — that's why CO is a fuel/pollutant). $\sum n_j{\bar{C}}_{p,j}=1(0.035)+2(0.045)+5.64(0.033)=0.035+0.090+0.18612=\mathrm{0.31112.}$ $T_{\text{ad}}=298+\frac{519.3}{0.31112}\approx 298+1669.1=1967\mathbf{K}.$ Lower than complete combustion (2339 K): incomplete burning wastes chemical energy.

First-pass rise: $T_{\text{ad}}-298=2041$. Scaling factor $f=1+0.20\cdot \frac{2041}{2000}=1+0.2041=1.2041$. New capacity: $0.39316\times 1.2041=0.47339$ kJ/K. $T_{\text{ad}}^{(1)}=298+\frac{802.3}{0.47339}\approx 298+1694.8=1993\mathbf{K}.$ The estimate dropped from 2339 K toward the realistic ~2230 K range — rising $C_p$ pulls AFT down. (More iterations would converge; here one pass shows the direction and magnitude.)

$\Delta H_{\text{rxn}}^{\circ }=[2(-393.5)+1(-241.8)]-[(+226.7)+0]=-1028.8-226.7=-1255.5\mathbf{kJ}.$ $\sum n_j{\bar{C}}_{p,j}=2(0.055)+1(0.045)=0.110+0.045=0.155\text{kJ/K}.$ $T_{\text{ad}}=298+\frac{1255.5}{0.155}\approx 298+8100.0=8398\mathbf{K}.$ Why so hot: the positive $\Delta H_f^{\circ }$ means acetylene is high on the energy ladder — it took energy to build, so it releases even more when it falls apart into stable $\mathrm{C}{\mathrm{O}}_2$ and ${\mathrm{H}}_2\mathrm{O}$. Subtracting a positive reactant enthalpy makes $\Delta H_{\text{rxn}}^{\circ }$ more negative. This model number is wildly high (dissociation caps real oxy-acetylene near ~3400 K), but it correctly explains why acetylene is a welding fuel of choice.

Reason 1 — material limits: no chamber wall survives 5000+ K; the effective usable temperature is set by cooling and metallurgy, not by the AFT ceiling. The N₂ "sponge" that lowers AFT also protects hardware. Reason 2 — dissociation losses: above ~2500 K products dissociate (${\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{OH}+\mathrm{H}$, $\mathrm{C}{\mathrm{O}}_2\rightleftharpoons \mathrm{CO}+\mathrm{O}$), absorbing energy so the real temperature never reaches the constant-$C_p$ estimate; you pay to make radicals that recombine downstream, wasting available thrust energy. So the 5831 K figure is a paper number, not a delivered one. (Bonus valid answers: carrying pure O₂ adds tank mass/cost; specific impulse depends on exhaust molar mass too, not temperature alone.)

The one-line recipe every problem used ::: BRSS — Balance, Reaction-enthalpy, Sensible-heat balance, Solve for $T_{\text{ad}}$. Sign rule in the master equation ::: $\sum n_j{\bar{C}}_p(T_{\text{ad}}-298)=-\Delta H_{\text{rxn}}^{\circ }$; since $\Delta H_{\text{rxn}}^{\circ }<0$ for combustion, the right side is positive. Why adding N₂ or excess air always lowers AFT ::: More product moles absorb heat (bigger denominator) with no extra chemical energy (same numerator). Why the constant-$C_p$ estimate is always an overestimate ::: It ignores rising $C_p$ and endothermic dissociation, both of which absorb energy and pull the real temperature down.