Question

Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30x10⁻⁸ at 700°C:2H₂S(g) ⇄ 2H₂(g) + S₂(g) If 0.45 mol of H₂S is placed in a 3.0-L container, what is the equilibrium concentration of H₂(g) at 700°C?

Answer 1

the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

0.00193 mol/L

Given that:

numbers of moles of H₂S = 0.59 moles

Volume = 3.0-L

Equilibrium constant  = 9.30 × 10⁻⁸

The equation for the reaction is given as :

2H₂S    ⇄   2H₂(g)  +  S₂(g)

The initial concentration of H₂S =

The initial concentration of H₂S =

= 0.1966 mol/L

The ICE table is shown be as :

                           2H₂S                ⇄         2H₂(g)        +        S₂(g)

Initial                    0.9166                           0                           0

Change                 -2 x                               +2 x                      + x

Equilibrium          (0.9166 - 2x)                   2x                         x

     

(since 2x < 0.1966 if solved through quadratic equation)

The equilibrium concentration for H₂(g) = 2x

∴

= 0.00193 mol/L

Thus, the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

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