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Consider the following equilibrium reaction at a given temperature: A (aq) + 3 B (aq) ⇌ C (aq) + 2 D (aq) Kc = 178. If initially the concentrations of the chemical species are as follows: [A]o = 0.364 M [B]o = 0.170 M [C]o = 0.132 M [D]o = 0.932 M to which direction will the reaction run to reach equilibrium?

Respuesta :

Answer: The reaction proceed toward product side or in forward direction.

Explanation:

[tex]K_c[/tex] is the constant of a certain reaction at equilibrium while [tex]Q_c[/tex] is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

[tex]A(aq.)+3B(aq.)\rightleftharpoons C(aq.)+2D(aq.)[/tex]

The expression of [tex]Q_c[/tex] for above equation follows:

[tex]Q_c=\frac{[D]^2[C]}{[A][B}^3}[/tex]

We are given:

[tex][A]_o=0.364M[/tex]

[tex][B]_o=0.170M[/tex]

[tex][C]_o=0.132M[/tex]

[tex][D]_o=0.932M[/tex]

Putting values in above equation, we get:

[tex]Q_c=\frac{(0.932)^2\times 0.132}{0.364\times (0.170)^3}=64.12[/tex]

We are given:

[tex]K_c=178[/tex]

There are 3 conditions:

  • When [tex]K_{c}>Q_c[/tex]; the reaction is product favored.
  • When [tex]K_{c}<Q_c[/tex]; the reaction is reactant favored.
  • When [tex]K_{c}=Q_c[/tex]; the reaction is in equilibrium.

As, [tex]K_c>Q_c[/tex], the reaction will be favoring product side.

Hence, the reaction proceed toward product side or in forward direction.

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