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.
Show Hint
Use the adiabatic energy balance |ΔH| = (total product plus inert moles) × Cp × (T minus 298), where N2 equals 4 times the O2 used.
Step 1: Understanding the Question:
We must find the flame temperature reached when methane burns completely in stoichiometric air (not pure oxygen), given the reaction enthalpy and a constant $C_p$ for every species involved.
Step 2: Key Formula or Approach:
Since air is only 20% oxygen by volume, we first find the total moles of air needed, then split off the nitrogen that tags along and treat it as extra mass that also has to be heated. The adiabatic condition means
\[ -\Delta H = n_{total}\,C_p\,\Delta T \]
Step 3: Detailed Explanation:
The reaction $CH_4 + 2O_2 \to CO_2 + 2H_2O$ needs 2 mol of $O_2$ per mole of methane. Since air is 20% $O_2$ by volume,
\[ \text{total air} = \frac{2}{0.20} = 10 \text{ mol} \]
Of this, the nitrogen fraction is 80%, so
\[ N_2 = 0.80\times 10 = 8 \text{ mol} \]
This matches the 4:1 ratio of $N_2$ to $O_2$ in the air, confirming the setup. The gas leaving the flame carries the products $CO_2$ (1 mol) and $H_2O$ (2 mol), plus the inert $N_2$ (8 mol), giving
\[ n_{total} = 1+2+8 = 11 \text{ mol} \]
All of them share $C_p = 50$ J/mol-K. Setting the heat released equal to the sensible heat gained,
\[ 850000 = 11(50)(T-298) \]
\[ T-298 = 1545.45 \]
\[ T = 1843.5 \text{ K} \]
Step 4: Final Answer:
The adiabatic flame temperature comes out close to 1843.5 K.
\[ \boxed{T \approx 1843.5 \text{ K}} \]
