Calculate the enthalpy of formation of ethene C2H4 from the given information
C2H4+H2O------ C2H5OH DH=-43kJmol
Enthalpy of formation of water And ethanol are -285.6kJmol and 52.3kJmol respectively
dHrxn = (n*dHf products) - (n*dHf reactants)
-43 = (1*52.3) - [(1*-285.6) + (dHf C2H4)]
Solve for dHf C2H4. Post your work if you get stuck.
To calculate the enthalpy of formation of ethene (C2H4), we need to first write the balanced chemical equation for the reaction given:
C2H4 + H2O -> C2H5OH
Given the enthalpy change (DH) for this reaction is -43 kJ/mol, and the enthalpies of formation of water (-285.6 kJ/mol) and ethanol (52.3 kJ/mol), we can use these values to calculate the enthalpy of formation of ethene.
The enthalpy change in the reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
DH = ∑Hf(products) - ∑Hf(reactants)
Let's calculate:
DH = (Hf(C2H5OH) + Hf(H2O)) - Hf(C2H4)
Given:
Hf(H2O) = -285.6 kJ/mol
Hf(C2H5OH) = 52.3 kJ/mol
Let's substitute the values into the equation:
-43 kJ/mol = (52.3 kJ/mol + (-285.6 kJ/mol)) - Hf(C2H4)
Simplifying the equation:
-43 kJ/mol = -233.3 kJ/mol - Hf(C2H4)
Rearranging the equation to solve for Hf(C2H4):
Hf(C2H4) = -233.3 kJ/mol + 43 kJ/mol
Hf(C2H4) = -190.3 kJ/mol
Therefore, the enthalpy of formation of ethene (C2H4) is -190.3 kJ/mol.
To calculate the enthalpy of formation of ethene (C2H4) using the given information, we can utilize Hess's Law. Hess's Law states that the overall enthalpy change of a chemical reaction is independent of the pathway taken.
We are given the equation:
C2H4 + H2O → C2H5OH ΔH = -43 kJ/mol (1)
We also know the enthalpy of formation of water (-285.6 kJ/mol) and ethanol (52.3 kJ/mol).
The enthalpy of formation for a compound, represented as ΔHf, is the enthalpy change for the formation of one mole of a compound from its constituent elements in their standard states.
By looking at equation (1), we can imagine that ethene (C2H4) is formed from ethene oxide (C2H5OH) and water (H2O). We can reverse this reaction to find the enthalpy change for the formation of ethene.
The given equation (1) can be reversed as follows:
C2H5OH → C2H4 + H2O ΔH = +43 kJ/mol (2)
The enthalpy change for equation (2) is the reverse of equation (1) in terms of its sign, so we change the sign of the value. Now, we need to account for the enthalpy change of water (ΔHf of H2O) and ethanol (ΔHf of C2H5OH).
The equation for the enthalpy change of formation of ethene (C2H4) can be written as:
C2H4 → C2H5OH - H2O ΔHf of C2H4?
Now, we can calculate the enthalpy of formation of ethene (ΔHf of C2H4) using the given values:
ΔHf of C2H4 = ΔHf of C2H5OH - ΔHf of H2O
Substituting the given values:
ΔHf of C2H4 = (52.3 kJ/mol) - (-285.6 kJ/mol)
Simplifying the expression:
ΔHf of C2H4 = 337.9 kJ/mol
Therefore, the enthalpy of formation of ethene (C2H4) is 337.9 kJ/mol.