Question

Calculate the standard enthalpy of formation of C₂H₄(g) from the following thermochemical equation

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g);
∆rH°=-1323Kj.
Given that ∆fH  CO₂(g) and H₂O(g) are -393.5 and -249 kJ mol⁻¹ respectively.​

Answer 1

To calculate the standard enthalpy of formation of C2H4(g), using Hess's Law, subtract the enthalpy of the reaction from the sum of the standard enthalpies of formation of the products. The result is 52.5 kJ/mol.

Step-by-step explanation:

The student is asking how to calculate the standard enthalpy of formation (ΔfH°) of ethylene (C2H4(g)) given the enthalpy of the reaction (ΔrH°) and the standard enthalpies of formation of carbon dioxide (CO2(g)) and water vapor (H2O(g)). To find the ΔfH° of C2H4(g), we can use Hess's Law which states that the total enthalpy change for a reaction is the same, regardless of the number of steps the reaction is carried out in. In this case, we know the enthalpy of the given reaction (C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g); ΔrH° = -1323 kJ) and the ΔfH° values for the products (CO2(g) = -393.5 kJ/mol, H2O(g) = -249 kJ/mol).

To calculate the ΔfH° of ethylene, we use the equation:

ΔfH° (ethylene) = ΔfH° (products) - ΔrH° (reaction)

Where

• ΔfH° (products) = 2(-393.5 kJ/mol) + 2(-249 kJ/mol)
• ΔrH° (reaction) = -1323 kJ

By substituting the values into the equation we get:

ΔfH° (ethylene) = (2 * -393.5 kJ/mol + 2 * -249 kJ/mol) + 1323 kJ

After calculation, ΔfH° (ethylene) is found to be 52.5 kJ/mol