Answer:
[HI] = 3.55 M
[H₂] = [I₂] = 0.425 M
Explanation:
The strategy here is to calculate Kc given the concentrations at equilibrium.Then we will setup an equilibrium expression in terms of the new concentrations and the amount that will be consumed of HI to reach equilibrium again.
2 HI (g) ⇆ H₂(g) + I₂(g)
Kc = [H₂][I₂]/[HI]² = 0.325 x 0.325 / 2.75² = 0.014
To account for the new equilibrium after adding 1.oo mol HI lets use the following table to make it easy for us.
mol HI H₂ I₂
Initially 3.75 0.325 0.325
Change -2x + x + x
Equilibrium 3.75 - 2x 0.325 + x 0.325 + x
These quantities at equilibrium has to obey the expression for Kc previously calculated. (Note we do not have to concern ourselves with calculating concentrations or switching to moles since the volume is 1 and M= mol/L)
We obtain the following equation:
(0.325 + x)² / (3.75 - 2x)² = 0.014
Taking square roots to both sides of this equation
( 0.325 + x )² / (3.75 - 2x) = 0.118
Solving for x we get the value of x = 0.10
Therefore the new equilibrium concentrations will be
[HI] = (3.75 - 0.20) M = 3.55 M
[H₂] = [I₂] = (0.325 + 0.10) M = 0.425 M
This answer can be checked by placing these values into the Kc. When we do it we get 0.014, so the answer is correct.
