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.

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

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 () for the given chemical reaction, is given by the equation:

At the original equilibrium state:

Therefore, at the new equilibrium state:

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

i think its 100.

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