# Step-by-Step Solution: Theoretical Flame Temperature for CO Combustion Let's solve **theoretical flame temperature** for the adiabatic combustion of CO with 100% excess air, **using Excel for calculations**. **Basis:** 1 mol CO (g). **Reactants enter at 100°C (373 K), at 1 atm.** --- ## **Step 1: Write the Combustion Reaction** \[ \text{CO (g)} + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \] --- ## **Step 2: Determine Moles of Reactants and Products** ### **A. Stoichiometric Oxygen Required** - 1 mol CO needs 0.5 mol O₂. ### **B. Excess Air (100%)** - 100% excess means **double the stoichiometric O₂** supplied. - Total O₂ supplied = \(0.5 \times 2 = 1.0 \) mol O₂ ### **C. Air Composition** - Air is 21% O₂ and 79% N₂ by mole. - Air needed: \( \frac{1.0~\text{mol O}_2}{0.21} = 4.76~\text{mol air} \) - N₂ supplied: \( 4.76~\text{mol air} \times 0.79 = 3.76~\text{mol N}_2 \) ### **D. Reactants Entering** - CO: 1 mol - O₂: 1 mol (from air) - N₂: 3.76 mol (from air) ### **E. Products After Combustion** - 1 mol CO₂ (from 1 mol CO) - O₂ leftover: \( 1.0 - 0.5 = 0.5 \) mol O₂ (from excess air) - N₂: 3.76 mol (inert, does not react) --- ## **Step 3: Write the Energy Balance (Adiabatic Reactor)** \[ \text{Total enthalpy in at 100}^\circ\text{C} = \text{Total enthalpy out at }T_f \] \[ \sum n_{in} [\Delta H^\circ_{f,25^\circ\text{C}} + \text{sensible } H_{100-25^\circ\text{C}}] = \sum n_{out} [\Delta H^\circ_{f,25^\circ\text{C}} + \text{sensible } H_{T-25^\circ\text{C}}] \] - **Adiabatic:** No heat loss to surroundings. --- ## **Step 4: Data Needed** | Species | ΔHf° (kJ/mol) | Cp (J/mol·K) | |---------|---------------|--------------| | CO (g) | -110.5 | ~29 | | O₂ (g) | 0 | ~29 | | N₂ (g) | 0 | ~29 | | CO₂ (g) | -393.5 | ~37 | > Values of Cp are **approximated**. For greater accuracy, use temperature-dependent Cp or tabulated enthalpy data. --- ## **Step 5: Calculate Enthalpy of Reactants at 100°C (373 K)** \[ \Delta H_{reactants} = \sum n_i \left[\Delta H^\circ_{f,25^\circ\text{C}} + \int_{25^\circ\text{C}}^{100^\circ\text{C}} C_{p,i} dT\right] \] **Cp is assumed constant for simplicity.** ### **A. CO (g):** \[ \Delta H_{CO} = (-110.5~\text{kJ/mol}) + (29~\text{J/mol·K}) \times (373-298)~\text{K} \times 1~\text{mol} \] \[ = -110.5 + (29 \times 75 / 1000) = -110.5 + 2.18 = -108.32~\text{kJ} \] ### **B. O₂ (g):** \[ \Delta H_{O_2} = 0 + (29 \times 75 / 1000) = 2.18~\text{kJ} \times 1~\text{mol} = 2.18~\text{kJ} \] ### **C. N₂ (g):** \[ \Delta H_{N_2} = 0 + (29 \times 75 / 1000) = 2.18~\text{kJ} \times 3.76~\text{mol} = 8.19~\text{kJ} \] ### **D. Total Reactant Enthalpy** \[ H_{reactants} = (-108.32) + (2.18) + (8.19) = -97.95~\text{kJ} \] --- ## **Step 6: Calculate Enthalpy of Products at Flame Temperature \(T_f\) (unknown)** ### **Products:** - CO₂: 1 mol - O₂: 0.5 mol - N₂: 3.76 mol \[ H_{products} = [\Delta H^\circ_{f,CO_2} + Cp_{CO_2}(T_f-298)] \times 1 \] \[ + [\Delta H^\circ_{f,O_2} + Cp_{O_2}(T_f-298)] \times 0.5 \] \[ + [\Delta H^\circ_{f,N_2} + Cp_{N_2}(T_f-298)] \times 3.76 \] \[ = [-393.5 + 37(T_f-298)/1000] \times 1 \] \[ + [0 + 29(T_f-298)/1000] \times 0.5 \] \[ + [0 + 29(T_f-298)/1000] \times 3.76 \] \[ H_{products} = -393.5 + 0.037(T_f-298) + 0.0145(T_f-298) + 0.109(T_f-298) \] \[ = -393.5 + [0.037 + 0.0145 + 0.109](T_f-298) \] \[ = -393.5 + 0.1605(T_f-298) \] --- ## **Step 7: Set Up the Energy Balance Equation** \[ H_{reactants} = H_{products} \] \[ -97.95 = -393.5 + 0.1605(T_f-298) \] \[ 0.1605(T_f-298) = -97.95 + 393.5 = 295.55 \] \[ T_f-298 = \frac{295.55}{0.1605} = 1841~\text{K} \] \[ T_f = 1841 + 298 = \boxed{2139~\text{K}} \] --- ## **Step 8: How to Set Up in Excel** ### **Excel Steps:** 1. **List all given data** (ΔHf°, Cp, moles) in one table. 2. **Calculate reactant enthalpy** at 100°C. 3. **Set up the enthalpy equation for products** as a function of \(T_f\). 4. **Use the "Goal Seek" tool** to solve for \(T_f\) that balances reactant and product enthalpy. ### **Excel Table Example:** | Species | n (mol) | ΔHf° (kJ/mol) | Cp (J/mol·K) | Initial T (K) | Final T (K) | Sensible ΔH (kJ) | Total ΔH (kJ) | |---------|---------|---------------|--------------|---------------|-------------|------------------|---------------| | CO | 1 | -110.5 | 29 | 373 | - | =29*(373-298)/1000 | -110.5 + [calc]| | O2 | 1 | 0 | 29 | 373 | - | =29*(373-298)/1000 | 0 + [calc] | | N2 | 3.76 | 0 | 29 | 373 | - | =29*(373-298)/1000 | 0 + [calc]*3.76| | Species | n (mol) | ΔHf° (kJ/mol) | Cp (J/mol·K) | Final T (K) | Sensible ΔH (kJ) | Total ΔH (kJ) | |---------|---------|---------------|--------------|-------------|------------------|---------------| | CO2 | 1 | -393.5 | 37 | T_f | =37*(T_f-298)/1000 | -393.5 + [calc] | | O2 | 0.5 | 0 | 29 | T_f | =29*(T_f-298)/1000 | [calc]*0.5 | | N2 | 3.76 | 0 | 29 | T_f | =29*(T_f-298)/1000 | [calc]*3.76 | - Sum reactant and product enthalpies. - In a cell, set `Total Reactant Enthalpy = Total Product Enthalpy`. - Use **Goal Seek** to adjust \(T_f\) until both sides are equal. --- ## **Final Answer** \[ \boxed{T_f \approx 2139~\text{K}} \] **Note:** For more accuracy, use tabulated enthalpy values for each gas at each temperature, especially at high temperatures (Cp is not truly constant). But this approach gives a good engineering estimate and is easily implemented in Excel.

# Step-by-Step Solution: Theoretical Flame Temperature for CO Combustion Let's solve **theoretical flame temperature** for the adiabatic combustion of CO with 100% excess air, using Excel for calculations. **Basis:** 1 mol CO (g). **Reactants enter at 100°C (373 K), at 1 atm.** --- ## Step 1: Write the Combustion Reaction \[ \text{CO (g)} + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \] --- ## Step 2: Determine Moles of Reactants and Products ### A. Stoichiometric Oxygen Required - 1 mol CO needs 0.5 mol O₂. ### B. Excess Air (100%) - Total O₂ supplied = \(0.5 \times 2 = 1.0 \) mol O₂ ### C. Air Composition - Air is 21% O₂ and 79% N₂ by mole. - Air needed: \( \frac{1.0~\text{mol O}_2}{0.21} = 4.76~\text{mol air} \) - N₂ supplied: \( 4.76~\text{mol air} \times 0.79 = 3.76~\text{mol N}_2 \) ### D. Reactants Entering - CO: 1 mol - O₂: 1 mol (from air) - N₂: 3.76 mol (from air) ### E. Products After Combustion - 1 mol CO₂ (from 1 mol CO) - O₂ leftover: \( 1.0 - 0.5 = 0.5 \) mol O₂ (from excess air) - N₂: 3.76 mol (inert, does not react) --- ## Step 3: Write the Energy Balance (Adiabatic Reactor) \[ \text{Total enthalpy in at 100}^\circ\text{C} = \text{Total enthalpy out at }T_f \] \[ \sum n_{in} [\Delta H^\circ_{f,25^\circ\text{C}} + \text{sensible } H_{100-25^\circ\text{C}}] = \sum n_{out} [\Delta H^\circ_{f,25^\circ\text{C}} + \text{sensible } H_{T-25^\circ\text{C}}] \] --- ## Step 4: Data Needed | Species | ΔHf° (kJ/mol) | Cp (J/mol·K) | |---------|---------------|---------------| | CO (g) | -110.5 | ~29 | | O₂ (g) | 0 | ~29 | | N₂ (g) | 0 | ~29 | | CO₂ (g) | -393.5 | ~37 | > Values of Cp are approximated. For greater accuracy, use temperature-dependent Cp or tabulated enthalpy data. --- ## Step 5: Calculate Enthalpy of Reactants at 100°C (373 K) \[ \Delta H_{reactants} = \sum n_i \left[\Delta H^\circ_{f,25^\circ\text{C}} + \int_{25^\circ\text{C}}^{100^\circ\text{C}} C_{p,i} dT\right] \] ### A. CO (g): \[ \Delta H_{CO} = (-110.5~\text{kJ/mol}) + (29~\text{J/mol·K}) \times (373-298)~\text{K} \times 1~\text{mol} \] \[ = -110.5 + (29 \times 75 / 1000) = -110.5 + 2.18 = -108.32~\text{kJ} \] ### B. O₂ (g): \[ \Delta H_{O_2} = 0 + (29 \times 75 / 1000) = 2.18~\text{kJ} \times 1~\text{mol} = 2.18~\text{kJ} \] ### C. N₂ (g): \[ \Delta H_{N_2} = 0 + (29 \times 75 / 1000) = 2.18~\text{kJ} \times 3.76~\text{mol} = 8.19~\text{kJ} \] ### D. Total Reactant Enthalpy \[ H_{reactants} = (-108.32) + (2.18) + (8.19) = -97.95~\text{kJ} \] --- ## Step 6: Calculate Enthalpy of Products at Flame Temperature \(T_f\) (unknown) ### Products: - CO₂: 1 mol - O₂: 0.5 mol - N₂: 3.76 mol \[ H_{products} = [\Delta H^\circ_{f,CO_2} + Cp_{CO_2}(T_f-298)] \times 1 \] \[ + [\Delta H^\circ_{f,O_2} + Cp_{O_2}(T_f-298)] \times 0.5 \] \[ + [\Delta H^\circ_{f,N_2} + Cp_{N_2}(T_f-298)] \times 3.76 \] \[ = [-393.5 + 37(T_f-298)/1000] \times 1 \] \[ + [0 + 29(T_f-298)/1000] \times 0.5 \] \[ + [0 + 29(T_f-298)/1000] \times 3.76 \] \[ H_{products} = -393.5 + 0.037(T_f-298) + 0.0145(T_f-298) + 0.109(T_f-298) \] \[ = -393.5 + [0.037 + 0.0145 + 0.109](T_f-298) \] \[ = -393.5 + 0.1605(T_f-298) \] --- ## Step 7: Set Up the Energy Balance Equation \[ H_{reactants} = H_{products} \] \[ -97.95 = -393.5 + 0.1605(T_f-298) \] \[ 0.1605(T_f-298) = -97.95 + 393.5 = 295.55 \] \[ T_f-298 = \frac{295.55}{0.1605} = 1841~\text{K} \] \[ T_f = 1841 + 298 = \boxed{2139~\text{K}} \] --- ## Step 8: How to Set Up in Excel ### Excel Steps: 1. **List all given data** (ΔHf°, Cp, moles) in one table. 2. **Calculate reactant enthalpy** at 100°C. 3. **Set up the enthalpy equation for products** as a function of \(T_f\). 4. **Use the "Goal Seek" tool** to solve for \(T_f\) that balances reactant and product enthalpy. ### Excel Table Example: | Species | n (mol) | ΔHf° (kJ/mol) | Cp (J/mol·K) | Initial T (K) | Final T (K) | Sensible ΔH (kJ) | Total ΔH (kJ) | |---------|---------|---------------|--------------|---------------|-------------|------------------|---------------| | CO | 1 | -110.5 | 29 | 373 | - | =29*(373-298)/1000 | -110.5 + [calc]| | O₂ | 1 | 0 | 29 | 373 | - | =29*(373-298)/1000 | 0 + [calc] | | N₂ | 3.76 | 0 | 29 | 373 | - | =29*(373-298)/1000 | 0 + [calc]*3.76| | Species | n (mol) | ΔHf° (kJ/mol) | Cp (J/mol·K) | Final T (K) | Sensible ΔH (kJ) | Total ΔH (kJ) | |---------|---------|---------------|--------------|-------------|------------------|---------------| | CO₂ | 1 | -393.5 | 37 | T_f | =37*(T_f-298)/1000 | -393.5 + [calc] | | O₂ | 0.5 | 0 | 29 | T_f | =29*(T_f-298)/1000 | [calc]*0.5 | | N₂ | 3.76 | 0 | 29 | T_f | =29*(T_f-298)/1000 | [calc]*3.76 | - Sum reactant and product enthalpies. - In a cell, set `Total Reactant Enthalpy = Total Product Enthalpy`. - Use **Goal Seek** to adjust \(T_f\) until both sides are equal. --- ## Final Answer \[ \boxed{T_f \approx 2139~\text{K}} \] **Note:** For more accuracy, use tabulated enthalpy values for each gas at each temperature, especially at high temperatures (Cp is not truly constant). But this approach provides a good engineering estimate and is easily implemented in Excel.