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A mixture of 9.22 moles of A, 10.11 moles of B, and 27.83 moles of C is placed in a one-liter
container at a certain temperature. The reaction is allowed to reach equilibrium. At equilibrium, the number of moles of B is 18.32. Calculate the equilibrium constant for the reaction:
A (g) + 2B (g) ⇌ 3C (g)

Respuesta :

pstnonsonjoku

The equilibrium constant is 1.3 considering the reaction as written in the question.

Equilibrium in chemical reactions

In a chemical reaction, the equilibrium constant is calculated based on the equilibrium concentration of each specie. The equation of this reaction is;

A (g) + 2B (g) ⇌ 3C (g).

The initial concentration of each specie is;

  • A - 9.22 M
  • B - 10.11 M
  • C - 27.83 M

The equilibrium concentration of B is 18.32 M

We now have to set up the ICE table as follows;

         A (g) +    2B (g) ⇌     3C (g)

I       9.22        10.11           27.83

C     -x              -x                +x

E   9.22 - x       10.11 - x       27.83 + x

The equilibrium concentration of B is 18.32 M hence;

10.11 - x = 18.32

x = 10.11 - 18.32 = -8.21

Hence;

Equilibrium concentration of A = 9.22 - (-8.21) = 17.43

Equilibrium concentration of C = 27.83 + (-8.21) = 19.62

Equilibrium constant K = [19.62]^3/[17.43] [18.32]^2

K = 1.3

Learn more about equilibrium constant: https://brainly.com/question/17960050

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