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
<em>(a) The cylinder should appear (color or colorless)</em>
At equilibrium, there is a mixture of A and B, so the cylinder should appear colored.
<em>(b) When the system is cooled, the cylinder's appearance becomes very light colored. Therefore, the reaction must be (endothermic or exothermic)</em>
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.
<em>Suppose the reaction equation were written as follows: 2 B(g) ⇄ A(g) </em>
<em>(c) Which of these statements would then be true?</em>
<em>a The value of K would not change.</em> FALSE. The new K would be the inverse of the direct K.
<em>b The ΔH value would have the same magnitude value but opposite sign. </em>TRUE. This is stated by Lavoisier-Laplace Law.
<em>c The K expression would be inverted.</em> TRUE. What was product before now is reactant and vice-versa.
<em>d The color of the cylinder would be darker.</em> FALSE. Changing the way the reaction is expressed has no effect on the equilibrium.
