Answer:
The equilibrium concentration of I₂ is 0.400 M and HI is 1.20 M, the Keq will be 0.112.
Step-by-step explanation:
Based on the given information, the equilibrium reaction will be,
2HI (g) ⇔ H₂ (g) + I₂ (g)
It is given that 4.00 mol of HI was filled in a flask of 2.00 L, thus, the concentration of HI will be,
= 4.00 mol/2.00 L
= 2.00 mol/L
Based on the reaction, the initial concentration of 2HI is 2.00, H₂ is 0 and I₂ is O. The change in the concentration of 2HI is -x, H₂ is x and I₂ is x. The equilibrium concentration of 2HI will be 0.200-x, H₂ is x and I₂ is x.
It is given that at equilibrium, the concentration of H₂ or x is 0.400 M.
Now the equilibrium concentration of HI will be,
= 2.00 -2x
= 2.00 - 2 × 0.400
= 1.20 M
The equilibrium concentration of I₂ will be,
I₂ = x
= 0.400 M
The equilibrium constant (Keq) will be,
Keq = [H₂] [I₂] / [HI]²
= (0.400) (0.400) / (1.20)²
= 0.112
Thus, the Keq of the reaction will be 0.112.