At a certain temperature, only 50% of HI is dissociated into H₂ and I₂.

2HI⇌ H₂ + I₂

Let's assume we start with 'a' moles of HI.
At equilibrium:
HI dissociated = 0.5a
HI remaining = a - 0.5a = 0.5a
H₂ formed = 0.5a / 2 = 0.25a
I₂ formed = 0.5a / 2 = 0.25a

Therefore, the equilibrium concentrations are:
[HI] = 0.5a
[H₂] = 0.25a
[I₂] = 0.25a