Let's solve the problem **step by step**: --- ## **Problem Restatement** **Methane (CH₄) is burned with 300 percent excess air in adiabatic constant pressure combustion. Initially, air and methane are at 1 atm and 298 K. Assuming complete combustion, determine the final temperature.** --- ## **Step 1: **Write the Combustion Reaction** For **complete combustion** of methane: \[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \] ### **Air Requirement** - **Stoichiometric O₂ needed:** 2 mol per 1 mol CH₄ - **Air contains 21% O₂ by mole:** - For 2 mol O₂, we need \( \frac{2}{.21} = 9.52 \) mol air ### **Excess Air** - **300% excess air** means \( 100\% + 300\% = 400\% \) of theoretical air. - **Total air supplied:** \( 4 \times 9.52 = 38.1 \) mol air per 1 mol CH₄ --- ## **Step 2: **Find the Product Composition ### **Oxygen Used:** - O₂ supplied: \( 38.1 \text{ mol air} \times .21 = 8. \) mol O₂ - O₂ consumed: 2 mol O₂ (for 1 mol CH₄) - **O₂ remaining (excess):** \( 8. - 2. = 6. \) mol ### **Nitrogen Supplied (from air):** - Air is 79% N₂ (by mole): \( 38.1 \text{ mol air} \times .79 = 30.1 \) mol N₂ ### **Final Mixture (per 1 mol CH₄):** - 1 mol CO₂ - 2 mol H₂O - 6 mol O₂ (excess) - 30.1 mol N₂ --- ## **Step 3: **Apply Energy Balance (Adiabatic System) **Adiabatic → No heat loss, so enthalpy in = enthalpy out** \[ \sum n_{\text{reactants}} h_{\text{reactants}} = \sum n_{\text{products}} h_{\text{products}} \] All **reactants start at 298 K**. \[ n_{\text{CH}_4} h^\circ_{\text{CH}_4} + n_{\text{O}_2} h^\circ_{\text{O}_2} + n_{\text{N}_2} h^\circ_{\text{N}_2} = \sum n_{\text{prod}} h_{\text{prod}}(T_f) \] But since \( h^\circ \) at 298 K is zero for elements, and H_f° for CH₄ and H₂O/CO₂ are known. --- ## **Step 4: **Plug Values and Solve ### **Reactant Side (at 298 K):** - 1 mol CH₄: \( \Delta H_f^\circ = -74.8 \) kJ/mol - 8 mol O₂, 30.1 mol N₂ (elements, so zero enthalpy at 298 K) ### **Product Side (at T_f):** - 1 mol CO₂ (\( \Delta H_f^\circ = -393.5 \) kJ/mol) - 2 mol H₂O (\( \Delta H_f^\circ = -241.8 \) kJ/mol, as vapor) - 6 mol O₂, 30.1 mol N₂: as gases \[ \text{Total enthalpy change from 298 K to } T_f: \sum n_i \int_{298}^{T_f} C_{p,i} dT \] ### **Energy Balance Equation** \[ \text{(Sum of product enthalpies at } T_f) = \text{(Sum of reactant enthalpies at 298 K)} \] \[ [1 \times \Delta H_{f,CO_2}^\circ + 2 \times \Delta H_{f,H_2O}^\circ + \sum n_i \int_{298}^{T_f} C_{p,i} dT] = [1 \times \Delta H_{f,CH_4}^\circ] \] \[ [(-393.5) + 2 \times (-241.8) + \text{Sensible heat of products}] = -74.8 \] \[ \text{Sensible heat of products} = -74.8 - [(-393.5) + 2 \times (-241.8)] \] \[ = -74.8 + 393.5 + 483.6 = 802.3 \text{ kJ} \] --- ## **Step 5: **Sensible Heat of Products Let’s denote: \[ Q = \sum n_i \int_{298}^{T_f} C_{p,i} dT = 802.3 \text{ kJ} \] Where: - \( n_{CO_2} = 1 \) - \( n_{H_2O} = 2 \) - \( n_{O_2} = 6 \) - \( n_{N_2} = 30.1 \) Average \( C_p \) values (approximate, in kJ/mol·K): - \( C_{p,CO_2} \approx .044 \) - \( C_{p,H_2O(g)} \approx .036 \) - \( C_{p,O_2} \approx .029 \) - \( C_{p,N_2} \approx .029 \) \[ Q = [1 \times .044 + 2 \times .036 + 6 \times .029 + 30.1 \times .029] \times (T_f - 298) \] \[ = [.044 + .072 + .174 + .8729] \times (T_f - 298) \] \[ = 1.1629 \times (T_f - 298) \] \[ 802.3 = 1.1629 \times (T_f - 298) \] \[ T_f - 298 = \frac{802.3}{1.1629} = 690 \text{ K} \] \[ T_f = 298 + 690 = 988 \text{ K} \] --- ## **Final Answer** ### **The final adiabatic flame temperature is:** \[ \boxed{988~\text{K}} \] --- ### **Summary Table** | Step | Description | Value/Equation | |------|-----------------------------------|---------------------------| | 1 | Stoichiometric air per 1 mol CH₄ | 9.52 mol air | | 2 | Supplied air (with 300% excess) | 38.1 mol air | | 3 | Products per 1 mol CH₄ | 1 CO₂, 2 H₂O, 6 O₂, 30.1 N₂ | | 4 | Energy balance | Sensible heat = 802.3 kJ | | 5 | Final temperature | 988 K | --- **If you need a more precise answer using temperature-dependent Cp values, let me know!**