Calculate ΔH for the process: Hg₂Cl₂(s) ⟶ 2Hg(l) + Cl₂(g) from the following information:

Question

Hg(l) + Cl₂(g)⟶HgCl₂(s) ΔH = −224 kJ

skyisabel1953

Answer 1

Given the equation:

Hg(l) + Cl₂(g) ⟶ HgCl₂(s) ΔH = -224 kJ

We can rearrange the given equation to find the enthalpy change for the process:

HgCl₂(s) ⟶ Hg(l) + Cl₂(g)

Since the enthalpy change for the reverse reaction is equal in magnitude but opposite in sign, we have:

ΔH = 224 kJ

Therefore, the enthalpy change for the process:

Hg₂Cl₂(s) ⟶ 2Hg(l) + Cl₂(g)

is 224 kJ.