Calcium chloride 6-hydrate, \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\), dehydrates according to the equation \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CaCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{P}=5.09 \times 10^{-44} \mathrm{at} 25^{\circ} \mathrm{C}\) What is the pressure of water vapor at equilibrium with a mixture of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CaCl}_{2}\) at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The pressure of water vapor at equilibrium with a mixture of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CaCl}_{2}\) at \(25^{\thinspace \circ} \mathrm{C}\) is the sixth root of \(5.09 \times 10^{-44}\).

Step by step solution

01

Write Expression For The Equilibrium Constant

The expression for the equilibrium constant (\( K_P \)) in terms of partial pressures for the dehydrating reaction of calcium chloride hexahydrate is given by \[ K_P = \left( P_{\mathrm{H_2O}} \right)^6 \] since the reaction involves the production of 6 moles of water vapor for each mole of calcium chloride hexahydrate that reacts, and there are no other gaseous products or reactants.
02

Solve For The Partial Pressure Of Water

Since no other pressures are involved, the sixth root of the equilibrium constant will give the pressure of water vapor: \[ P_{\mathrm{H_2O}} = \sqrt[6]{K_P} \] Now we can input the given value of the equilibrium constant into the equation.
03

Calculate The Pressure Of Water Vapor

Plugging the given equilibrium constant into the equation, we get: \[ P_{\mathrm{H_2O}} = \sqrt[6]{5.09 \times 10^{-44}} \] Use a calculator to find the sixth root of the equilibrium constant.
04

Present Final Answer

After calculating the sixth root of the equilibrium constant, we arrive at the pressure of water vapor at equilibrium.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical system where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of reactants and products over time. In the given exercise, we encounter an equilibrium involving calcium chloride hexahydrate dehydrating to form anhydrous calcium chloride and water vapor.

At equilibrium, the system has reached a steady state where the dehydration of calcium chloride hexahydrate and the potential rehydration from water vapor and calcium chloride occur at the same rate. The equilibrium constant, denoted as \( K_P \) when expressed in terms of partial pressures of gases, quantifies the ratio of the product of the partial pressures of the products to the reactants, raised to the power of their stoichiometric coefficients. This constant is intrinsic to the reaction at a given temperature.

Understanding chemical equilibrium is crucial because it allows us to predict the concentrations of substances involved in a reaction at any given point. Moreover, manipulating the conditions of the reaction can shift the equilibrium position, influencing the degree of conversion and yield of the desired products.
Partial Pressure
Partial pressure is a concept used to describe the pressure that a single component of a gas mixture would exert if it occupied the entire volume of the mixture at the same temperature. In relation to the dehydrating reaction of calcium chloride hexahydrate, the partial pressure of water vapor (\( P_{\text{H}_2\text{O}} \)) is the pressure contributed by just the water vapor in the equilibrium mixture.

In solving the problem, we consider the partial pressure of water vapor only because it is the sole gas in the reaction. Understanding the concept of partial pressure is essential because it helps to relate the physical behavior of gases to the chemical reaction taking place. For gases, the equilibrium expression in terms of partial pressure reflects the impact of each component's pressure on the whole system, enabling us to solve for unknown pressures if the equilibrium constant is known.

Furthermore, in real-world applications, chemists use the notion of partial pressure to control reactions involving gases, ensuring that the desired reaction proceeds with the highest efficiency possible.
Le Chatelier's Principle
Le Chatelier's principle is a fundamental concept that predicts the response of a system in chemical equilibrium upon changes in concentration, temperature, volume, or pressure. It states that if an external change is applied to a system at equilibrium, the system will adjust itself in such a way as to counteract that change.

In the exercise's context, if the temperature were to change, for example, Le Chatelier's principle would allow us to predict the direction in which the equilibrium would shift. Since we know the dehydration reaction is endothermic (absorbs heat), an increase in temperature would favor the forward reaction, producing more water vapor and shifting the equilibrium towards the products.

Understanding Le Chatelier's principle also enables one to manipulate conditions to maximize the yield of reactions. For chemists, this theory is a tool for predicting how changes in conditions will impact the extent of reaction, allowing for more efficient industrial processes and the ability to steer reactions toward a desired outcome. The principle serves as a guide for determining optimal reaction conditions in various chemical industries.

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Most popular questions from this chapter

A reaction is represented by this equation: \(2 \mathrm{W}(a q) \rightleftharpoons \mathrm{X}(a q)+2 \mathrm{Y}(a q) \quad K_{c}=5 \times 10^{-4}\) (a) Write the mathematical expression for the equilibrium constant. (b) Using concentrations of \(\leq 1 M,\) identify two sets of concentrations that describe a mixture of \(\mathrm{W}, \mathrm{X},\) and \(\mathrm{Y}\) at equilibrium.

A solution is saturated with silver sulfate and contains excess solid silver sulfate: \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) A small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive \(\mathrm{Ag}^{+}\) ions. Explain this observation.

Write the mathematical expression for the reaction quotient, \(Q_{c}\), for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)=2 \mathrm{NO}_{2}(g)\) (d) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{NH}_{4} \mathrm{Cl}(s)=\mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) (f) \(2 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightleftharpoons 2 \mathrm{PbO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) (g) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)=2 \mathrm{H}_{2} \mathrm{O}(l)\) (h) \(S_{8}(g) \rightleftharpoons 8 S(g)\)

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas. (a) Write the expression for the equilibrium constant \(\left(K_{c}\right)\) for the reversible reaction \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=98.7 \mathrm{kJ}\) (b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added? (c) What will happen to the concentration of each reactant and product at equilibrium if \(\mathrm{H}_{2} \mathrm{O}\) is removed? (d) What will happen to the concentration of each reactant and product at equilibrium if \(\mathrm{H}_{2}\) is added? (e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased? (f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be \(100 \%\) Is \(K_{c} > 1, < 1,\) or \(\approx 1\) for a titration reaction?

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