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

Short Answer

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The indicator will be Pink if rain is imminent, due to the equilibrium shift favoring the formation of cobalt(II) chloride hexahydrate (CoCl₂·6H₂O) in response to increased water vapor concentration.

Step by step solution

01

Determine the color of the indicator

Due to the equilibrium shift favoring the formation of cobalt(II) chloride hexahydrate (CoCl₂·6H₂O), the color of the indicator will be Pink, which is the color of CoCl₂·6H₂O. So, the indicator will be Pink if rain is imminent.

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Key Concepts

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

Chemical Equilibrium
Understanding the dynamics of chemical reactions is fundamental to the study of chemistry, and chemical equilibrium plays a key role in this. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, meaning the reactants and products are formed at the same rate. In our exercise, we have the equilibrium between anhydrous cobalt(II) chloride and its hydrated form, cobalt(II) chloride hexahydrate.

The reaction can be represented as: \[\text{CoCl}_2(s) + 6\text{H}_2\text{O}(g) \rightleftharpoons \text{CoCl}_2 \cdot 6\text{H}_2\text{O}(s)\]. In the presence of moisture (like high humidity before rain), the reaction shifts to the right, forming more of the pink hydrate. Conversely, in dry conditions, the equilibrium shifts to the left, and the anhydrous form, which has a purple color, predominates. Through Le Chatelier's Principle, we understand that a change in conditions leads to an equilibrium shift to counteract that change. This shift is what causes the color change in our hygroscopic weather predictor.
Hygroscopic Substances
Hygroscopic substances have a knack for attracting and holding water molecules from the surrounding environment through absorption or adsorption. Cobalt(II) chloride is a particularly well-known hygroscopic material that undergoes a noticeable physical change upon water absorption.

The reaction, \[\text{CoCl}_2(s) + 6\text{H}_2\text{O}(g) \rightleftharpoons \text{CoCl}_2 \cdot 6\text{H}_2\text{O}(s)\], showcases cobalt(II) chloride's hygroscopic nature as it bonds with water to form its hydrate. This property is leveraged in applications such as moisture indicators in novelty devices, desiccants for keeping goods dry, and even in laboratories to control humidity levels. Their ability to react with water vapor and change color based on the degree of hydration or dehydration makes them invaluable tools.
Predicting Weather Changes
While modern meteorology uses complex instruments and computer models to predict the weather, simple chemical principles can still provide clues about upcoming changes. Substances like cobalt(II) chloride serve as rudimentary weather prediction tools. The color change from purple to pink in the presence of moisture is a visual cue of increased humidity, which often precedes rain.

Novelty weather-indicating devices capitalize on this chemistry to provide a handy, albeit basic, forecast tool. As moisture in the air increases, these devices, containing cobalt(II) chloride, absorb water and transition to a pink state. Such rudimentary methods can be surprisingly accurate in the short term and provide a fun way to engage with the principles of science and nature.

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Most popular questions from this chapter

Explain the difference between \(K, K_{\mathrm{p}},\) and \(Q\).

Consider the reaction $$\mathbf{P}_{4}(g) \longrightarrow 2 \mathbf{P}_{2}(g)$$, where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{K},\) the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium.

In a study of the reaction $$3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)$$,at \(1200 \mathrm{K}\) it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at \(1200 \mathrm{K}\). (Hint: Apply Dalton's law of partial pressures.)

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 mol/L. Calculate the value of \(K\) for this reaction.

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g):\).$$\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g)$$.A 2.00 -g sample of XY (molar mass \(=165 \mathrm{g} / \mathrm{mol}\) ) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\). The pressure is held constant at 0.967 atm. As XY begins to dissociate, the piston moves until 35.0 mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\).

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