A mixture initially contains A, B, and C in the following concentrations: [A] = 0.300 M , [B] = 1.05 M , and [C] = 0.550 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.140 M and [C] = 0.710 M . Calculate the value of the equilibrium constant, Kc.

Respuesta :

Answer:

Kc = 9.52.

Explanation:

  • The equilibrium system:

A + 2B ⇌ C,

Kc = [C]/[A][B]²,

Concentration:     [A]                [B]              [C]

At start:               0.3 M         1.05 M        0.55 M

At equilibrium:   0.3 - x        1.05 - 2x     0.55 + x

                            0.14 M        1.05 - 2x      0.71 M

  • For the concentration of [A]:

∵ 0.3 M - x = 0.14 M.

∴ x = 0.3 M - 0.14 M = 0.16 M.

∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.

∵ Kc = [C]/[A][B]²

∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.

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