Adiabatic flame temperature (idealized):\nMethane combusts with the stoichiometric amount of air: CH4 + 2 O2 → CO2 + 2 H2O (all species in the gas phase). Given ΔH°298 = −730 kJ per mole CH4 and an average Cp of 40 J·mol^−1·K^−1 for all gases, estimate the maximum temperature rise (°C) of the products.

Introduction / Context:
The adiabatic flame temperature (or maximum temperature rise) is a cornerstone estimate in combustion design. It assumes that all heat released by the reaction raises the sensible enthalpy of the products without losses, dissociation, or radiation. This problem requests an approximate ΔT using a constant average heat capacity for simplicity.

Given Data / Assumptions:

• Reaction: CH4 + 2 O2 → CO2 + 2 H2O (g).
• ΔH°298 = −730 kJ per mol CH4 (heat released).
• Stoichiometric air: O2 provided by air; N2 accompanies O2 with N2/O2 ≈ 79/21.
• Average Cp for all gases: 40 J·mol^−1·K^−1 = 0.040 kJ·mol^−1·K^−1.
• Adiabatic, no dissociation, constant Cp approximation.

Concept / Approach:
Energy balance for an adiabatic reactor: total heat released by reaction equals sensible heat gain of products. Compute total moles of gaseous products, multiply by average Cp, and solve for ΔT = (−ΔH_reaction) / (Σ n_p Cp,avg). Temperature rise in °C equals rise in K for differences.

Step-by-Step Solution:

Verification / Alternative check:
The result lies near textbook adiabatic flame temperatures for methane–air when using low average Cp and ignoring dissociation; more detailed models (temperature-dependent Cp, H2O condensation at low T, dissociation) reduce the estimate.

Why Other Options Are Wrong:

• 1225, 1335, 1525 °C underestimate ΔT given the stated −730 kJ/mol and low Cp assumption.

Common Pitfalls:
Forgetting to include nitrogen from air in heat capacity; mixing units (J vs kJ); using liquid-water heat of reaction instead of all-vapor basis.

Final Answer:
1735