Question

At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106 M; [I2] = 0.022 M; [HI] = 1.29 M Calculate the new equilibrium concentration of HI (in M) if the equilibrium concentrations of H2 and I2 are 0.95 M and 0.019 M respectively.

Answer 1

The new equilibrium concentration of HI: [HI] = 3.589 M

Step-by-step explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (
`K_(c)`) for the given chemical reaction, is given by the equation:

`K_(c) = \frac {[HI]^(2)}{[H_(2)]\: [I_(2)]}`

At the original equilibrium state:

`K_(c) = \frac {(1.29\: M)^(2)}{(0.106\: M) * (0.022\: M)}`

`K_(c) = \frac {1.6641}{0.002332} = 713.59`

Therefore, at the new equilibrium state:

`K_(c) = \frac {[HI]^(2)}{(0.95\: M) * (0.019\: M)}`

`\Rightarrow K_(c) = 713.59 = \frac {[HI]^(2)}{0.01805}`

`\Rightarrow [HI]^(2) = 713.59 * 0.01805 = 12.88`

`\Rightarrow [HI] = \sqrt {12.88} = 3.589 M`

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M