What is the enthalpy for the following reaction?
overall: C2H4+H2O--->C2H5OH
is -1411kJ correct??
DeltaH= -40 kJ/mol
2CO2 + 3H2O→C2H5OH + 3O2, ΔH=1370 kJ/mol
+ C2H4 + 3O2→2CO2 + 2H2O, ΔH=−1410 kJ/mol
gives you
C2H4 + H2O→C2H5OH, ΔH=−40 kJ/mol
or is 44kJ correct?
bqbbalabak
That was all the question gave me I can not figure out how to get the correct answer.
I need answers
To determine the enthalpy change for a reaction, you typically need the enthalpy values of the reactants and products. In this case, we need the enthalpy of C2H4 (ethylene), H2O (water), and C2H5OH (ethanol).
To obtain accurate values, you can refer to a reliable source such as a chemistry reference book or an online database. These sources provide standard enthalpy of formation (\(ΔH_f^{\circ}\)) values for various compounds.
The standard enthalpy of formation is the change in enthalpy that accompanies the formation of one mole of a compound from its elements in their standard states (usually solid or liquid). The standard state of an element is the form it exists in under standard conditions of 298 K (25°C) and 1 atmosphere pressure.
Once you have the \(ΔH_f^{\circ}\) values for C2H4, H2O, and C2H5OH, you can calculate the enthalpy change for the reaction by subtracting the sum of the \(ΔH_f^{\circ}\) values of the reactants from the sum of the \(ΔH_f^{\circ}\) values of the products. The equation would look like this:
Enthalpy change (\(ΔH\)) = Σ\(ΔH_f^{\circ}\) (products) - Σ\(ΔH_f^{\circ}\) (reactants)
Please note that the \(ΔH_f^{\circ}\) values are typically given per mole of substance. Therefore, make sure to take into account the stoichiometric coefficients of the reactants and products when performing the calculations.
Without specific values for the \(ΔH_f^{\circ}\) of C2H4, H2O, and C2H5OH, it is not possible to confirm if -1411 kJ is the correct enthalpy change for the given reaction.