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Step 1:I'll solve these reactions using the standard enthalpies of formation method.
General Formula: $$\Delta H^{\circ}_{\text{reaction}} = \sum n \Delta H^{\circ}_f(\text{products}) - \sum m \Delta H^{\circ}_f(\text{reactants})
I'll break down each reaction step-by-step.
Step 2:: \mathrm{NaOH}(\mathrm{s}) + \mathrm{HCl}(\mathrm{g}) \rightarrow \mathrm{NaCl}(\mathrm{s}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{g})
= -133.8 \text{ kJ/mol}
Step 3:: 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})
= -566.0 \text{ kJ/mol}
Step 4:: \mathrm{CH}_{4}(\mathrm{g}) + 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{l})
= -890.3 \text{ kJ/mol}
Step 5:: 2 \mathrm{H}_{2}\mathrm{S}(\mathrm{g}) + 3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 2 \mathrm{SO}_{2}(\mathrm{g})
= -1123.6 \text{ kJ/mol}
Step 6:: 2 \mathrm{NO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})
= -113.0 \text{ kJ/mol}
Final Answer
1. \Delta H^{\circ} = - 133.8 \text{ kJ/mol} 2. \Delta H^{\circ} = - 566.0 \text{ kJ/mol} 3. \Delta H^{\circ} = - 890.3 \text{ kJ/mol} 4. \Delta H^{\circ} = - 1123.6 \text{ kJ/mol} 5. \Delta H^{\circ} = - 113.0 \text{ kJ/mol}
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