Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium: $$ \underset{\text { Purple }}{\mathrm{CoCl}_{2}(s)}+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \underset{\text { Pink }}{\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)} $$ What color will such an indicator be if rain is imminent?

Short Answer

Expert verified
The indicator will be pink when rain is imminent, signaling an increase in humidity.

Step by step solution

01

Understand the effect of increased humidity on the equilibrium

When humidity increases, more water molecules are present in the air. According to Le Chatelier's principle, when a system at equilibrium is subject to a change, the system will shift the equilibrium to counteract the change. In this case, the equilibrium system will shift in a direction that consumes more of the added species (water molecules) to re-establish equilibrium.
02

Identify the direction of the equilibrium shift

As we have an increased amount of water molecules in a humid environment, according to Le Chatelier's principle, the equilibrium will shift towards the right to consume more water molecules. This means more cobalt(II) chloride hydrate (CoCl2·6H2O) will be formed.
03

Determine the color of the indicator

Since the equilibrium shifts to the right and more cobalt(II) chloride hydrate is formed, the color of the indicator will be closer to the color of the CoCl2·6H2O. This compound has a pink color when solid. Therefore, the rain indicator will be pink if rain is imminent. So, the indicator will be pink when rain is imminent, signaling an increase in humidity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Understanding chemical equilibrium is essential for grasping how reactions behave over time. Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. Imagine a busy city intersection where cars are entering and leaving at the same rate; the number of cars within the intersection area stays consistent.

In the context of a chemical reaction, this does not mean the reactants and products are in equal concentrations, but that their rates of formation are balanced. The position of the equilibrium gives us information about the proportions of substances in the mixture at equilibrium.
Humidity and Chemical Equilibrium
Humidity plays a significant role in influencing chemical equilibria, especially in reactions involving hydrates and water vapor. Le Chatelier's principle helps us understand this influence by stating that when a system in equilibrium is disturbed, it will adjust in a way that counteracts the disturbance.

When humidity increases, the air contains more water vapor. A system containing a hydrate will respond by shifting the equilibrium to use up excess water vapor. This can be observed in everyday phenomena such as the pink coloration of a humidity indicator, where the formation of a cobalt(II) chloride hydrate intensifies with higher humidity levels, indicating a shift in the chemical equilibrium towards the hydrated form.
Cobalt(II) Chloride Hydrate
Cobalt(II) chloride hydrate, with its distinct color changes, serves as an excellent example of a chemical hydrate and its equilibrium with water vapor. This hydrate specifically transitions between a purple anhydrous form, CoCl2, and a pink hydrated form, CoCl2·6H2O, in response to atmospheric humidity.

The vivid color change associated with cobalt(II) chloride provides a visual indicator of the equilibrium shift. When the hydrate is exposed to dry conditions, the pink color fades to purple as water molecules are released into the environment. Conversely, in humid conditions, the compound takes on a pink hue due to the formation of CoCl2·6H2O, revealing how humidity can directly impact a chemical's properties.

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Most popular questions from this chapter

Consider the following reactions. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) and \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is \(1.5\) atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\). a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\). b. If the volume of the cylinder is increased until the total pressure is \(1.0 \mathrm{~atm}\) (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\). c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{~atm}\) )?

Given the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\), consider the following situations: i. You have \(1.3 M \mathrm{~A}\) and \(0.8 \mathrm{M} \mathrm{B}\) initially. ii. You have \(1.3 M \mathrm{~A}, 0.8 M \mathrm{~B}\), and \(0.2 M \mathrm{C}\) initially. iii. You have \(2.0 \mathrm{M} \mathrm{A}\) and \(0.8 \mathrm{M} \mathrm{B}\) initially. Order the preceding situations in terms of increasing equilibrium concentration of D. Explain your order. Then give the order in terms of increasing equilibrium concentration of \(\mathrm{B}\) and explain.

A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{~K}\), where the following reactions occurred: $$ \begin{aligned} 2 \mathrm{FeSO}_{4}(s) & \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) & \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{aligned} $$ After equilibrium was reached, the total pressure was \(0.836\) atm and the partial pressure of oxygen was \(0.0275\) atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a \(0.256\) -mol sample of naphthalene has a mass of \(32.8 \mathrm{~g}\). What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene \((s) \rightleftharpoons\) naphthalene \((g)\) \(K=4.29 \times 10^{-6}(\) at \(298 \mathrm{~K})\) If \(3.00 \mathrm{~g}\) solid naphthalene is placed into an enclosed space with a volume of \(5.00 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\), what percentage of the naphthalene will have sublimed once equilibrium has been established?

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