Given the reaction: C_2H_4 + H_2O → C_2H_5OH, what is the enthalpy change for this reaction in terms of standard enthalpies of formation (ΔH_f^ext{°})?

• A. ΔH = ΔH_f^ext{°}(C_2H_5OH) + ΔH_f^ext{°}(C_2H_4) + ΔH_f^ext{°}(H_2O)
• B. ΔH = ΔH_f^ext{°}(C_2H_4) + ΔH_f^ext{°}(H_2O) - ΔH_f^ext{°}(C_2H_5OH)
• C. ΔH = ΔH_f^ext{°}(C_2H_5OH) - [ΔH_f^ext{°}(C_2H_4) + ΔH_f^ext{°}(H_2O)]
• D. ΔH = ΔH_f^ext{°}(C_2H_4) - [ΔH_f^ext{°}(C_2H_5OH) + ΔH_f^ext{°}(H_2O)]

Answer: (C)

Verified step by step guidance

1. Identify the reaction and the substances involved: ethylene (\(C_2H_4\)), water (\(H_2O\)), and ethanol (\(C_2H_5OH\)). The reaction is \(C_2H_4 + H_2O \rightarrow C_2H_5OH\).
2. Recall that the enthalpy change of a reaction (\(\Delta H\)) can be calculated using the standard enthalpies of formation (\(\Delta H_f^{\circ}\)) of the products and reactants with the formula: \(\Delta H = \sum \Delta H_f^{\circ} (\text{products}) - \sum \Delta H_f^{\circ} (\text{reactants})\)
3. Write the expression for this specific reaction by substituting the compounds: \(\Delta H = \Delta H_f^{\circ}(C_2H_5OH) - [\Delta H_f^{\circ}(C_2H_4) + \Delta H_f^{\circ}(H_2O)]\)
4. Understand that the enthalpy of formation values represent the energy change when one mole of a compound is formed from its elements in their standard states, so this formula accounts for the energy difference between products and reactants.
5. Use the correct values of \(\Delta H_f^{\circ}\) from a data table to plug into the formula and calculate the enthalpy change for the reaction.