Question

In the presence of excess of anhydrous \(\mathrm{SrCl}_{2}\), the amount of water taken up is governed by \(K_{p}=10^{12} \mathrm{~atm}^{-4}\) for the following reaction at \(273 \mathrm{~K}\) $$ \mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{SrCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) $$ What is equilibrium vapour pressure (in torr) of water in a closed vessel that contains \(\mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) ?\) (a) \(0.001\) torr (b) \(10^{3}\) torr (c) \(0.76\) torr (d) \(1.31\) torr

Short answer

The equilibrium vapour pressure of water is 760,000 torr, which is not among the given options, indicating a possible error in the question or answer choices.

Step-by-step solution

- Understand the Chemical Equilibrium

In this step, recognize that the given reaction is in a state of dynamic equilibrium. In a closed container, SrCl2·2H2O solid reacts with water vapor to form SrCl2·6H2O solid. Since SrCl2 is in excess, its concentration does not affect the equilibrium. The equilibrium constant expression for this reaction in terms of partial pressure (Kp) is given by the equation Kp = ([H2O(g)])^4, where [H2O(g)] is the equilibrium partial pressure of water vapor.

- Setting up the Equilibrium Expression

Using the given value of Kp, which is 10^12 atm^-4, we set up the equilibrium expression. According to the given reaction, the stoichiometry indicates that 4 moles of water vapor react to form one mole of the hexahydrate solid. Since we are interested in the vapor pressure of water, the expression is Kp = (P_H2O)^4.

- Solving for Equilibrium Vapour Pressure

We now solve for the equilibrium water vapor pressure (P_H2O) by taking the fourth root of both sides of the equation: (P_H2O)^4 = 10^12, which results in P_H2O = 10^3 atm. To express the pressure in torr instead of atmospheres, we use the conversion factor that 1 atm = 760 torr. Therefore, P_H2O = 10^3 atm * 760 torr/atm.

- Convert the Pressure into Torr

Carry out the conversion by multiplying the pressure in atmospheres by the conversion factor to get the pressure in torr: P_H2O(torr) = P_H2O(atm) * 760 torr/atm.

- Final Calculation

The final step is to multiply 10^3 by 760 to find the equilibrium vapor pressure in torr: P_H2O(torr) = 10^3 * 760 torr.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when a reversible reaction reaches a state where the rates of the forward and reverse reactions are equal, leading to no net change in the concentrations of reactants and products over time. In a closed system, the substances involved in the reaction can change their forms but their total mass remains constant. At equilibrium, these substances are present in a particular ratio that is maintained unless the system is disturbed by external factors like temperature or pressure changes.

Understanding chemical equilibrium is crucial when analyzing reactions such as the hydration of anhydrous SrCl₂, a process that can reach equilibrium under specific conditions. In a classroom setting, visual aids or animations showing dynamic equilibrium can significantly enhance comprehension, demonstrating that while individual molecules continue to react, the overall system does not change.

Equilibrium Constant (Kp)

The equilibrium constant (Kp) is a value that expresses the relationship between the concentrations of the gaseous reactants and products at equilibrium. For reactions involving gases, Kp is defined based on the partial pressures of the reactants and products. It is temperature-dependent and remains constant unless the temperature changes. The expression for Kp is derived from the balanced chemical equation, using the partial pressures and stoichiometry of the reaction components.

For example, the Kp given for the SrCl₂ hydration reaction is extremely high (\(10^{12} \text{ atm}^{-4}\)), indicating that at equilibrium, the reaction strongly favors the formation of the hexahydrate product. Students can benefit from examples that illustrate how different values of Kp affect the position of equilibrium, as well as practice exercises that involve calculating Kp for various reactions.

Vapor Pressure Calculations

Vapor pressure is an essential concept when dealing with the physical properties of substances. It refers to the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. The equilibrium vapor pressure of a substance is an indication of its volatility. Calculating vapor pressure becomes relevant in determining the behavior of substances in a closed system, as in the hydration reaction of SrCl₂.

When a hydrate is in equilibrium with water vapor in a closed vessel, the vapor pressure can be calculated using the equilibrium constant, as demonstrated in the exercise. In a teaching scenario, incorporating practical demonstrations of vapor pressure, like observing water boiling at lower temperatures under reduced pressure, can solidify the concept for students.

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction, derived from the balanced chemical equation. It allows chemists to predict the amounts of substances consumed and formed in a reaction. Stoichiometry extends to reactions at equilibrium, informing the formulation of the equilibrium expression based on the mole ratio of reactants and products.

In the given exercise, the stoichiometry indicates a 4:1 mole ratio of water vapor to the hexahydrate product. Engaging students with stoichiometric calculations in diverse scenarios, including gaseous reactions and equilibria, can dramatically improve their problem-solving skills. Simple analogies, like recipes in cooking, can remarkably help students grasp the essence of stoichiometry and its practical applications.