What is the enthalpy for the following reaction?

C2H4+H2O--->C2H5OH

Look up delta H^o_f for products and reactants, then

delta Hrxn = (delta H sum products)-(delta H sum reactants)

formation enthalpy of ethane -> –83.8 kJ/mol

formation enthalpy of water -> −291.83 kJ/mol

formation enthalpy of ethanol -> –277.7 kJ/mol

I got 97.93 and got it wrong I only have one try left. I don't know whats wrong.

What phases are these reagents in? It makes a difference if they are in gas or liquid phase. The delta H is considerably different (although the differences may not be different).

To find the enthalpy (ΔH) for a reaction, you need to determine the difference in enthalpy between the products and reactants. The enthalpy change can be obtained from experimental data, such as from a calorimeter experiment, or by using tabulated values called standard enthalpies of formation.

To calculate the enthalpy change using standard enthalpies of formation, follow these steps:

1. Write the balanced chemical equation for the reaction:
C2H4 + H2O → C2H5OH

2. Determine the standard enthalpies of formation (∆Hf°) for each compound involved in the reaction:
C2H4: ? kJ/mol
H2O: -285.8 kJ/mol (from tables)
C2H5OH: ? kJ/mol

3. Add up the enthalpies of formation for the products, making sure to consider stoichiometric coefficients:
∆Hf°(C2H5OH) = (∆Hf°(C2H4) + ∆Hf°(H2O))

4. Subtract the sum of the enthalpies of formation for the reactants:
ΔH = ∆Hf°(C2H5OH) - (∆Hf°(C2H4) + ∆Hf°(H2O))

To get the specific values of the standard enthalpies of formation (∆Hf°), you can refer to a reliable source such as a chemistry textbook, database, or online resource.