What is the percentage dissociation of HI at 444°C, if the experiment is started with one mole of HI — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

For the dissociation reaction 2HI(g) ⇌ H₂(g) + I₂(g), the change in the number of moles of gas (Δn_g) is 2 - 2 = 0. The addition of an inert gas like He at constant volume has no effect on the equilibrium state for reactions where Δn_g = 0. Thus, the percentage dissociation remains unchanged at 22.22%. Therefore, correct answer is C.