Certainly! Let's go through each problem step-by-step clear explanations. --- ## Problem 6: \(\Delta H_{rxn) for overall combustion of propane Given Data: - Reactions: 1. \( \mathrm{C(s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO(g)} \quad \Delta H = -110.5\, \text{kJ/mol} \) 2. \( 3\, \mathrm{C(s)} + 4\, \mathrm{H_2(g)} \rightarrow \mathrm{C_3H_8(g)} \quad \Delta H = +241.8\, \text{kJ/mol} \) 3. \( \mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow \mathrm{H_2O(g)} \quad \Delta H = -241.8\, \text{kJ/mol} \) 4. \( \mathrm{C_3H_8(g)} + 5\, \mathrm{O_2(g)} \rightarrow 3\, \mathrm{CO_2(g)} + 4\, \mathrm{H_2O(g)} \quad \Delta H = -393.5\, \text{kJ/mol} \) ### Goal: Determine \(\Delta H_{rxn}\) for the overall combustion of propane (reaction 4). --- ### Step 1: Write the overall combustion of propane The overall combustion reaction: \[ \mathrm{C_3H_8(g)} + 5\, \mathrm{O_2(g)} \rightarrow 3\, \mathrm{CO_2(g)} + 4\, \mathrm{H_2O(g)} \] We need to find \(\Delta H\) for this. --- ### Step 2: Use Hess's Law Hess's Law states: \[ \Delta H_{overall} = \sum \text{(adjusted reactions' enthalpies)} \] Our approach: - Combine reactions 1-4, manipulating (multiplying, reversing) reactions to get the overall reaction. --- ### Step 3: Manipulate reactions - **Reaction 2** (formation of propane): Use as-is. - **Reaction 3** (formation of water): Use to cancel water. - **Reaction 4** (combustion of propane): Use as-is. - **Reaction 1** (formation of CO): Use to relate CO formation. --- ### Step 4: Write the steps to get the overall reaction **a. Formation of propane:** \[ 3\, \mathrm{C(s)} + 4\, \mathrm{H_2(g)} \rightarrow \mathrm{C_3H_8(g)} \quad \Delta H = +241.8\, \text{kJ} \] **b. Combustion of propane:** \[ \mathrm{C_3H_8(g)} + 5\, \mathrm{O_2(g)} \rightarrow 3\, \mathrm{CO_2(g)} + 4\, \mathrm{H_2O(g)} \quad \Delta H = -393.5\, \text{kJ} \] **c. Formation of water:** \[ \mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow \mathrm{H_2O(g)} \quad \Delta H = -241.8\, \text{kJ} \] --- ### Step 5: Combine reactions - To get the total combustion, we can think of: \[ \text{(Formation of propane)} + \text{(combustion of propane)} + \text{(formation of water)} \text{ (to cancel water)}. \] But since the reactions involve formation and combustion, the Hess's Law approach involves: - Reverse reaction 3 (to cancel water on both sides) - Sum all reactions, adjusting signs as needed. --- ### Step 6: Final calculation \[ \boxed{ \Delta H_{rxn} = \Delta H_2 + \Delta H_4 - 4 \times \Delta H_3 } \] Where: - \(\Delta H_2 = +241.8\, \text{kJ}\) - \(\Delta H_4 = -393.5\, \text{kJ}\) - \(\Delta H_3 = -241.8\, \text{kJ}\) Note: Water formation is used 4 times in the combustion, so: \[ \Delta H_{rxn} = 241.8 + (-393.5) - 4 \times (-241.8) \] \[ = 241.8 - 393.5 + 967.2 \] \[ = (241.8 - 393.5) + 967.2 = -151.7 + 967.2 = 815.5\, \text{kJ} \] --- ## **Final Answer for Problem 6:** \[ \boxed{ \Delta H_{rxn} \approx +815.5\, \text{kJ} } \] --- ## Problem 7: Use Hess's Law to determine \(\Delta H_{rxn}\) for overall reaction ### Given Data: - Reactions: 1. \( \mathrm{N_2(g)} + 3\, \mathrm{H_2(g)} \rightarrow 2\, \mathrm{NH_3(g)} \quad \Delta H = -46.2\, \text{kJ/mol} \) 2. \( 2\, \mathrm{H_2O(l)} \rightarrow 2\, \mathrm{H_2(g)} + \mathrm{O_2(g)} \quad \Delta H = 571.5\, \text{kJ} \) 3. \( 2\, \mathrm{HNO_3(aq)} \rightarrow \mathrm{N_2O_3(g)} + \mathrm{H_2O(l)} \quad \Delta H = -140.2\, \text{kJ} \) 4. \( \mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow \mathrm{H_2O(g)} \quad \Delta H = -241.8\, \text{kJ} \) 5. \( \mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)} \quad \Delta H = -44.1\, \text{kJ} \) ### Goal: Determine \(\Delta H_{rxn}\) for the overall reaction (not explicitly given), but based on the reactions, it seems to involve nitrogen, hydrogen, and possibly ammonia or related compounds. --- ### Step 1: Clarify the overall target reaction Given the reactions, a likely overall reaction might be: \[ \mathrm{N_2(g)} + 3\, \mathrm{H_2(g)} \rightarrow 2\, \mathrm{NH_3(g)} \] which is a typical Haber process. --- ### Step 2: Use Hess's Law **Note:** To determine \(\Delta H_{rxn}\) for the overall reaction, sum reactions to produce this target. --- ### Step 3: Express all reactions as needed - Reaction 1 is already the synthesis of ammonia. - Reaction 2 relates to water formation. - Reaction 3 involves nitrate formation. - Reactions 4 and 5 relate to water formation from hydrogen and oxygen. --- ### Step 4: Combine reactions Assuming the goal is to find \(\Delta H\) for: \[ \mathrm{N_2(g)} + 3\, \mathrm{H_2(g)} \rightarrow 2\, \mathrm{NH_3(g)} \] The key is: - Use Reaction 1 directly: \(\Delta H = -46.2\, \text{kJ/mol}\) - Reactions 2-5 can be combined to relate water and nitric acid reactions, possibly to cancel intermediates. --- ### **Without the explicit overall reaction provided,** the best approach is to recognize that **Reaction 1 directly gives the enthalpy change for ammonia synthesis**. ### **Final answer:** \[ \boxed{ \Delta H_{rxn} \approx -46.2\, \text{kJ/mol} } \] --- **Summary:** - For Problem 6, the overall combustion enthalpy of propane is approximately **+815.5 kJ**. - For Problem 7, based on the reactions, the enthalpy change for synthesizing ammonia is approximately **-46.2 kJ**. Let me know if you'd like further clarification or details!