Calculate the enthalpy of reaction for the combustion of ethene.
To calculate the enthalpy of reaction for the combustion of ethene, we need to know the balanced chemical equation for the combustion of ethene, as well as the standard enthalpy of formation values for the reactants and products involved. The balanced chemical equation for the combustion of ethene is:
C2H4 + 3O2 -> 2CO2 + 2H2O
Using the standard enthalpy of formation values, we can calculate the enthalpy of reaction using the formula:
ΔH = Σ(∆Hf(products)) - Σ(∆Hf(reactants))
The standard enthalpy of formation values for ethene (C2H4), carbon dioxide (CO2), and water (H2O) are -52.5 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.
Now, substituting these values into the formula:
ΔH = [2*(-393.5 kJ/mol) + 2*(-285.8 kJ/mol)] - [1*(-52.5 kJ/mol) + 3*(0 kJ/mol)]
Calculating this expression:
ΔH = [-787 kJ/mol] - [-52.5 kJ/mol]
ΔH = -787 kJ/mol + 52.5 kJ/mol
ΔH = -734.5 kJ/mol
Therefore, the enthalpy of reaction for the combustion of ethene is -734.5 kJ/mol.
To calculate the enthalpy of reaction for the combustion of ethene, we need to know the balanced chemical equation for the reaction and the standard enthalpy of formation values for all the reactants and products involved.
The balanced chemical equation for the combustion of ethene is:
C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(g)
Step 1: Determine the standard enthalpy of formation (ΔHf) values. These values represent the change in enthalpy when one mole of a compound is formed from its elements in their standard states (usually at 25°C and 1 atmosphere pressure).
ΔHf(C2H4) = +52.3 kJ/mol (obtained from reference sources)
ΔHf(O2) = 0 kJ/mol (elements in their standard states have no change in enthalpy)
ΔHf(CO2) = -393.5 kJ/mol (obtained from reference sources)
ΔHf(H2O) = -241.8 kJ/mol (obtained from reference sources)
Step 2: Use the balanced chemical equation to calculate the overall enthalpy change (ΔH) for the reaction. This can be done by summing up the product values and subtracting the sum of the reactant values, while considering the stoichiometric coefficients.
ΔH = (2 * ΔHf(CO2) + 2 * ΔHf(H2O)) - (ΔHf(C2H4) + 3 * ΔHf(O2))
ΔH = (2 * (-393.5 kJ/mol) + 2 * (-241.8 kJ/mol)) - (+52.3 kJ/mol + 3 * 0 kJ/mol)
ΔH = -1309 kJ/mol
Therefore, the enthalpy change (ΔH) for the combustion of ethene is -1309 kJ/mol. The negative sign indicates that the reaction releases heat (exothermic).
C2H4 + 3O2 ==> 2CO2 + 2H2O
dHrxn = (n*dHf products) - (n*dHf reactants).
You can look up the values for dHformation in your text or notes.