Question

Use Hess's Law to determine the enthalpy change associated with the following reaction: C₂H₄(g) + Cl₂(g) => C₂H₄Cl₂(g). Given:

(i) 4HCI(g) + O₂(g) => 2C1₂(g) + 2H₂O(l) + 202.4 kJ
(ii) 2HCl(g) + C₂H₄(g) + 1/2O₂(g) => C₂H₄Cl₂(g) + H₂O(l) + 320.8 kJ

Answer 1

To determine the enthalpy change for the reaction C₂H₄(g) + Cl₂(g) ⇒ C₂H₄Cl₂(g) using Hess's Law, we reverse and halve the first given equation, then subtract the second given equation to get an enthalpy change of +219.6 kJ.

Step-by-step explanation:

To use Hess's Law to determine the enthalpy change associated with the reaction C₂H₄(g) + Cl₂(g) ⇒ C₂H₄Cl₂(g), we need to manipulate the given equations to sum up to the target reaction. Given the reactions:

1. 4HCl(g) + O₂(g) ⇒ 2Cl₂(g) + 2H₂O(l) ▲H = +202.4 kJ
2. 2HCl(g) + C₂H₄(g) + ½O₂(g) ⇒ C₂H₄Cl₂(g) + H₂O(l) ▲H = +320.8 kJ

We first reverse reaction (i) in order to have Cl₂(g) on the reactant side and then divide it by 2 to be consistent with the target equation's stoichiometry:

2Cl₂(g) + 2H₂O(l) ⇒ 4HCl(g) + O₂(g) ▲H = -202.4 kJ / 2 = -101.2 kJ

Next, we subtract reaction (ii) from the reversed and halved reaction (i) to cancel out the HCl and O₂, leaving us with the desired reaction:

C₂H₄(g) + Cl₂(g) ⇒ C₂H₄Cl₂(g) ▲H = -101.2 kJ + 320.8 kJ = +219.6 kJ

The enthalpy change for the reaction C₂H₄(g) + Cl₂(g) ⇒ C₂H₄Cl₂(g) is +219.6 kJ.