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Hydrogen iodide decomposes according to the reaction2HI(g) ⇄ H₂(g) + I₂(g) A sealed 1.50-L container initially holds 0.00623 mol ofH₂, 0.00414 mol of I₂, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H₂(g) is 0.00467 M. What are the concentrations of HI(g) and I₂(g)?



Answer :

tutorconsortium344

The concentration of  and  at equilibrium is, 0.0158 M and 0.00302 M respectively.

First we have to calculate the concentration of

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                           

Initial conc.      0.0163     0.00415      0.00276

At eqm.        (0.0163-2x) (0.00415+x)  (0.00276+x)

As we are given:

Concentration of  at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of  at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of  at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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