For the following reaction between methane and stoichiometric air having composition 20 vol.% \(O_2\) and 80 vol.% \(N_2\), the estimated adiabatic flame temperature (rounded off to one decimal place) is ________ Kelvin.
\[
CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(g)
\]
Given: \(\Delta H = -850\) kJ/mol at 298 K. Assume the specific heat at constant pressure (\(C_p\)) for each reactant and product is 50 J/mol-K and is independent of temperature.
\[ CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(g) \]
Given: \(\Delta H = -850\) kJ/mol at 298 K. Assume the specific heat at constant pressure (\(C_p\)) for each reactant and product is 50 J/mol-K and is independent of temperature.
Show Hint
Correct Answer: 1842.5
Solution and Explanation
The fuel burns in stoichiometric air, not in pure oxygen. Air here is 20 vol.% \(O_2\) and 80 vol.% \(N_2\), so for every mole of \(O_2\) supplied there are 4 moles of \(N_2\) riding along as an inert diluent.
The balanced reaction is
\[ CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(g) \]
This needs 2 mol of \(O_2\) per mole of \(CH_4\), so the \(N_2\) that comes in with this air is
\[ N_2 = 4\times 2 = 8 \text{ mol} \]
Step 2: List every mole that leaves the flame front.
After combustion, the hot gas leaving the flame contains the products plus the untouched nitrogen. Per mole of \(CH_4\) burned:
\[ CO_2 = 1 \text{ mol}, \quad H_2O = 2 \text{ mol}, \quad N_2 = 8 \text{ mol (inert, unreacted)} \]
Total moles of hot gas:
\[ n_{total} = 1 + 2 + 8 = 11 \text{ mol} \]
Step 3: Set up the adiabatic energy balance.
In an adiabatic flame, no heat leaves the system. All the heat released by the reaction at 298 K goes into heating the product gas mixture, including the inert \(N_2\), from 298 K up to the flame temperature \(T\). Since every species shares the same \(C_p = 50\) J/mol-K, and it does not change with temperature,
\[ |\Delta H| = n_{total}\,C_p\,(T - 298) \]
Step 4: Substitute the numbers and solve for T.
\[ 850000 = 11\times 50\times(T-298) \]
\[ 850000 = 550\,(T-298) \]
\[ T - 298 = \frac{850000}{550} = 1545.5 \]
\[ T = 298 + 1545.5 = 1843.5 \text{ K} \]
Final Answer:
The adiabatic flame temperature works out to about 1843.5 K, which sits inside the accepted band of 1842.5 to 1844.5 K.
\[ \boxed{T \approx 1843.5 \text{ K}} \]
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