Let's break down the problem step by step, providing detailed explanation and calculation at each step. --- ### **1. Write the Reaction** \[ \text{Cyclohexane to Benzene:} \quad \ce{C6H12(g) -> C6H6(g) + 3H2(g)} \] --- ### **2. Find the Heat of Reaction at Standard Conditions (\( \Delta H^\circ_{rxn, 25^\circ C} \))** We need standard enthalpies of formation (\( \Delta H^\circ_f \)): | Substance | \( \Delta H^\circ_f \) (kJ/mol) | |------------------|-----------------------| | \(\ce{C6H12(g)}\) (cyclohexane) | -29.9 | | \(\ce{C6H6(g)}\) (benzene) | 82.9 | | \(\ce{H2(g)}\) | 0 | \[ \Delta H^\circ_{rxn} = [1 \times \Delta H^\circ_f(\ce{C6H6}) + 3 \times \Delta H^\circ_f(\ce{H2})] - [1 \times \Delta H^\circ_f(\ce{C6H12})] \] \[ = [1 \times 82.9 + 3 \times 0] - [1 \times (-29.9)] = (82.9 + 0) - (-29.9) = 82.9 + 29.9 = 112.8\, \text{kJ/mol} \] **So,** \[ \Delta H^\circ_{rxn, 25^\circ C} = +112.8\, \text{kJ/mol} \] --- ### **3. Reactor Stoichiometry and Flow** **Basis:** 1 mol of \(\ce{C6H12}\) fed - Conversion = 70% - Entering: 1 mol \(\ce{C6H12}\) + 0.5 mol \(\ce{N2}\) (inert) **Moles reacted:** \[ \text{Moles }\ce{C6H12}\text{ reacted} = 1 \times 0.70 = 0.70 \] **Moles unreacted:** \[ \text{C6H12 unreacted} = 1 - 0.70 = 0.30 \] **Moles of products:** - For each mole \(\ce{C6H12}\) reacted: 1 mol \(\ce{C6H6}\) + 3 mol \(\ce{H2}\) - \(\ce{C6H6}\): \(0.70\) mol - \(\ce{H2}\): \(3 \times 0.70 = 2.10\) mol **Exit Stream:** - \(\ce{C6H12}\) (unreacted): \(0.30\) mol - \(\ce{C6H6}\): \(0.70\) mol - \(\ce{H2}\): \(2.10\) mol - \(\ce{N2}\): \(0.5\) mol (inert) --- ### **4. Enthalpy Calculation Scheme** We need to find the **heat to be added or removed** (\(Q\)) to maintain energy balance. #### **General energy balance:** \[ Q = \text{(Total enthalpy of exit at T)} - \text{(Total enthalpy of feed at 50°C)} \] (Assume reactor at constant pressure, ignore kinetic/potential energy.) #### **Reference: 25°C (298 K)** **Step 1:** Bring all feed to 25°C (reference) - \(\ce{C6H12(g)}\) and \(\ce{N2(g)}\) from 50°C to 25°C **Step 2:** Reaction at 25°C (use \(\Delta H^\circ_{rxn}\)) **Step 3:** Heat products from 25°C to exit temperature (either 25°C or 300°C) --- ### **5. Calculate Heats of Sensible Enthalpy (\( \Delta H_{sensible} \))** We'll need molar heat capacities (\(C_p\)): | Substance | \(C_p\) (J/mol·K) | |------------------|------------------| | \(\ce{C6H12(g)}\) | 156 | | \(\ce{C6H6(g)}\) | 134 | | \(\ce{H2(g)}\) | 29 | | \(\ce{N2(g)}\) | 29 | --- ### **a. Exit at 25°C** **Step 1:** Cool feed from 50°C to 25°C - \(\Delta T = 25 - 50 = -25\) K \[ \Delta H_{feed} = [1 \times 156 + 0.5 \times 29] \times (-25) = [156 + 14.5] \times (-25) = 170.5 \times (-25) = -4262.5\, \text{J} = -4.26\, \text{kJ} \] **Step 2:** Reaction at 25°C - 0.70 mol reacted, \(\Delta H^\circ_{rxn} = 112.8\) kJ/mol \[ \Delta H_{rxn} = 0.70 \times 112.8 = 78.96\, \text{kJ} \] **Step 3:** Heat products from 25°C to 25°C (\(\Delta T = 0\)), so \(\Delta H_{products} = 0\) **Total Q:** \[ Q = \Delta H_{feed} + \Delta H_{rxn} + \Delta H_{products} \] \[ Q = (-4.26) + 78.96 + 0 = 74.7\, \text{kJ} \] **Interpretation:** Since Q is positive, **heat must be removed** from the reactor. --- ### **b. Exit at 300°C** **Step 1:** Cool feed from 50°C to 25°C (**same as above**): \(-4.26\) kJ **Step 2:** Reaction at 25°C (**same as above**): \(+78.96\) kJ **Step 3:** Heat products from 25°C to 300°C (\(\Delta T = 275\) K) \[ \Delta H_{products} = \sum n_i C_{p,i} \Delta T \] \[ = (0.30 \times 156 + 0.70 \times 134 + 2.10 \times 29 + 0.5 \times 29) \times 275 \] \[ = (46.8 + 93.8 + 60.9 + 14.5) \times 275 = (216.0) \times 275 = 59,400\, \text{J} = 59.4\, \text{kJ} \] **Total Q:** \[ Q = (-4.26) + 78.96 + 59.4 = 134.1\, \text{kJ} \] **Interpretation:** Again, **heat must be removed** from the reactor. --- ### **Final Answers (to 1 decimal place):** #### **a. Exit at 25°C:** > **Heat removed:** \(\boxed{74.7\ \text{kJ} \text{ per mol of } C_6H_{12} \text{ fed}}\) #### **b. Exit at 300°C:** > **Heat removed:** \(\boxed{134.1\ \text{kJ} \text{ per mol of } C_6H_{12} \text{ fed}}\) --- #### **Summary Table** | Case | Exit Temp | Heat Removed (\(Q\)) | |------|-----------|-----------------------| | (a) | 25°C | 74.7 kJ/mol | | (b) | 300°C | 134.1 kJ/mol | --- **Notes:** - Positive Q means heat must be **removed**. - All calculations based on 1 mole of cyclohexane fed. - If you need per mole of cyclohexane **reacted**, divide by 0.70. - If you want to include more precise heat capacity data or temperature dependence, use mean values or integrate \(C_p(T)\), but for most practical purposes, the above calculation suffices. --- **Let me know if you need the calculation for a different basis or more detailed steps!**

Let's carefully determine the heat of reaction and the heat exchange with the environment for the given process. --- ### Step 1: Write the reaction \[ \ce{C6H12(g) -> C6H6(g) + 3H2(g)} \] --- ### Step 2: Find the standard enthalpy of reaction at 25°C Using standard enthalpies of formation (\(\Delta H^\circ_f\)): | Substance | \(\Delta H^\circ_f\) (kJ/mol) | |-----------------------|------------------------------| | \(\ce{C6H12(g)}\) | -29.9 | | \(\ce{C6H6(g)}\) | 82.9 | | \(\ce{H2(g)}\) | 0 | Calculate: \[ \Delta H^\circ_{rxn} = [\Delta H^\circ_f(\ce{C6H6}) + 3 \times 0] - [\Delta H^\circ_f(\ce{C6H12})] = 82.9 - (-29.9) = 112.8\, \text{kJ/mol} \] **This is the enthalpy change at 25°C.** --- ### Step 3: Determine the molar flow rates Given: - Feed: 1 mol \(\ce{C6H12}\) plus inert \(0.5\) mol \(\ce{N2}\) - Conversion: 70% **Reacted:** \[ 0.70\, \text{mol} \quad \text{(reaction for 1 mol feed)} \] **Unreacted:** \[ 0.30\, \text{mol} \] **Products formed:** - \(\ce{C6H6}\): 0.70 mol - \(\ce{H2}\): \(3 \times 0.70 = 2.10\) mol **Inert \(\ce{N2}\):** remains at 0.5 mol --- ### Step 4: Energy considerations The total heat exchange involves: - Heating/cooling of inlet gases to the reference temperature (50°C) - Reaction enthalpy - Heating of products to exit temperature --- ### Step 5: Molar heat capacities (\(C_p\)) used: | Substance | \(C_p\) (J/mol·K) | |-----------------------|------------------| | \(\ce{C6H12(g)}\) | 156 | | \(\ce{C6H6(g)}\) | 134 | | \(\ce{H2(g)}\) | 28.8 | | \(\ce{N2(g)}\) | 29.1 | --- ### Step 6: Heating/cooling the inlet gases from 50°C to 25°C Temperature change: \[ \Delta T = 25 - 50 = -25\,^\circ C \] Heat: \[ Q_{\text{cooling}} = \left[ (1 \times 156) + (0.5 \times 29.1) \right] \times (-25\,K) \] \[ = (156 + 14.55) \times (-25) = 170.55 \times (-25) = -4263.75\, \text{J} \approx -4.26\, \text{kJ} \] *Negative indicates cooling (removing heat).* --- ### Step 7: Reaction enthalpy at 25°C \[ \Delta H_{reaction} = 0.70 \times 112.8\, \text{kJ} = 78.96\, \text{kJ} \] *This is the heat released (+) or absorbed (-) during the reaction.* --- ### Step 8: Heating the products from 25°C to exit temperature **a. If gases leave at 25°C:** \[ \Delta T = 0\,^\circ C \] No additional heating. **b. If gases leave at 300°C:** \[ \Delta T = 300 - 25 = 275\,^\circ C \] Calculate the heat needed: \[ Q_{\text{heating}} = \left[ (0.30 \times 156) + (0.70 \times 134) + (2.10 \times 28.8) + (0.5 \times 29.1) \right] \times 275 \] \[ = (46.8 + 93.8 + 60.5 + 14.55) \times 275 \] \[ = 215.65 \times 275 \approx 59,319\, \text{J} \approx 59.3\, \text{kJ} \] --- ### **Final calculations:** --- ### **a. Exit gases at 25°C** \[ Q_{\text{total}} = Q_{\text{cooling}} + \text{reaction enthalpy} + Q_{\text{heating}} \] \[ = -4.26\, \text{kJ} + 78.96\, \text{kJ} + 0\, \text{kJ} = \boxed{74.7\, \text{kJ}} \] *Heat must be removed (since net Q is positive, but the initial cooling is negative, overall energy balance indicates removal).* --- ### **b. Exit gases at 300°C** \[ Q_{\text{total}} = -4.26\, \text{kJ} + 78.96\, \text{kJ} + 59.3\, \text{kJ} = \boxed{134.0\, \text{kJ}} \] *Again, heat is removed from the reactor (Q positive).* --- ### **Final answers:** | Condition | Heat removed (per mol \(\ce{C6H12}\) fed) | |------------|----------------------------------| | (a) at 25°C | **74.7 kJ** | | (b) at 300°C | **134.0 kJ** | --- ### **Summary:** - The conversion releases a significant amount of heat (reaction enthalpy). - Additional heat must be removed to keep the reactor at the specified exit conditions. - The amount of heat removed increases with the exit temperature because more energy is needed to heat the products. --- **Let me know if you'd like further clarification or calculations based on different assumptions!**