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Two gases, A and B, are at equilibrium in a sealed cylinder. Individually, gas A is colorless, while gas B is dark colored. The reaction might be written as: A(g) equilibrium reaction arrow 2 B(g). (a) The cylinder should appear (color or colorless) (b) When the system is cooled, the cylinder's appearance becomes very light colored. Therefore, the reaction must be (endothermic or exothermic) Suppose the reaction equation were written as follows: 2 B(g) equilibrium reaction arrow A(g) (c) Which of these statements would then be true? (Select all that apply.) a The value of K would not change. b The ΔH value would have the same magnitude value but opposite sign. c The K expression would be inverted. d The color of the cylinder would be darker.

Respuesta :

dsdrajlin

Answer:

(a) color

(b) endothermic

(c)

b The ΔH value would have the same magnitude value but opposite sign.

c The K expression would be inverted.

Explanation:

Let's consider the following reaction at equilibrium.

      A(g) ⇄ 2 B(g)

colorless    dark colored

(a) The cylinder should appear (color or colorless)

At equilibrium, there is a mixture of A and B, so the cylinder should appear colored.

(b) When the system is cooled, the cylinder's appearance becomes very light colored. Therefore, the reaction must be (endothermic or exothermic)

According to Le Chatelier's Principle when a perturbation is made to a system at equilibrium it will react to counteract such effect. When the system is cooled, it will tend to increase the temperature by releasing heat. In this case, the reaction is endothermic so when the reverse reaction is favored, colorless A is favored as well.

Suppose the reaction equation were written as follows: 2 B(g) ⇄ A(g)

(c) Which of these statements would then be true?

a The value of K would not change. FALSE. The new K would be the inverse of the direct K.

b The ΔH value would have the same magnitude value but opposite sign. TRUE. This is stated by Lavoisier-Laplace Law.

c The K expression would be inverted. TRUE. What was product before now is reactant and vice-versa.

d The color of the cylinder would be darker. FALSE. Changing the way the reaction is expressed has no effect on the equilibrium.

pstnonsonjoku

The gas appears dark colored at equilibrium and the reaction is endothermic.

First of all, the equation of the reaction is; A(g) ⇄ 2B(g). We know that A is a colorless gas while B is a dark colored gas. It follows that, at equilibrium, the cylinder should appear dark colored.

When the cylinder appears light colored, then the equilibrium position has shifted to the left hand side according to the reaction equation. If increasing the temperature makes the gas to appear colorless then the reaction is endothermic. When the temperature of an endothermic reaction is decreased, the reaction proceeds in the reverse direction.

Two statements are true as written when the reaction is rewritten as  2B(g)  ⇄ A(g);

  • The ΔH value would have the same magnitude value but opposite sign.
  • The K expression would be inverted.

Note that, when the ΔH value of the reverse reaction has the magnitude as the  ΔH value of the forward reaction but carries an opposite sign. Similarly, the equilibrium constant of the reverse reaction is the inverse of the equilibrium constant of the forward reaction hence the K expression would be inverted.

Learn more: https://brainly.com/question/10038290

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