Question

\(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) \(+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), K_{\mathrm{e}}\) for this equilibrium is \(1.0 \times 10^{-4} \mathrm{~atm}^{2}\) at \(25^{\circ} \mathrm{C}\). What is the maximum pressure of water vapour (moisture) in the atmosphere, below which the pentahydrate is efflorescent? (a) \(7.60 \mathrm{~mm}\) (b) \(0.01 \mathrm{~mm}\) (c) \(0.076 \mathrm{~mm}\) (d) \(760 \mathrm{~mm}\)

Short answer

The maximum pressure of water vapour is \(0.01 \mathrm{~mm}\).

Step-by-step solution

Understand the Equilibrium Expression

Write the equilibrium expression for the given reaction. The reaction is: \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O(s)} \rightleftharpoons \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O(s)} + 2 \mathrm{H}_{2} \mathrm{O(g)}\). The equilibrium constant \(K_{e}\) only includes the concentrations (or partial pressures) of the gaseous species, thus \(K_{e}={[\mathrm{H}_{2} \mathrm{O(g)}]}^{2}\).

Set Up the Equilibrium Constant

Using the given equilibrium constant \(K_{e} = 1.0 \times 10^{-4} \mathrm{~atm}^{2}\), and the fact that \(K_{e}={[\mathrm{H}_{2} \mathrm{O(g)}]}^{2}\), we need to find the partial pressure of water vapor when the reaction is at equilibrium.

Calculate the Pressure of Water Vapor

As \(K_{e}={[\mathrm{H}_{2} \mathrm{O(g)}]}^{2}\), solving for \(\mathrm{H}_{2} \mathrm{O(g)}\) gives us \(\mathrm{H}_{2} \mathrm{O(g)} = \sqrt{K_{e}} = \sqrt{1.0 \times 10^{-4} \mathrm{~atm}^{2}}\). Calculate the square root to get the pressure in atm.

Convert the Pressure to mmHg

Use the conversion factor \(1 \mathrm{~atm} = 760 \mathrm{~mmHg}\) to convert the pressure from atm to mmHg to determine the maximum vapor pressure for efflorescence.

Key concepts

Equilibrium Constant Expression

When studying chemical reactions that reach a state where the reactants and products are formed at the same rate, we are dealing with a dynamic equilibrium. To understand this state quantitively, we use the equilibrium constant expression, symbolized by K or K_e. This numerical value indicates the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients, at equilibrium.
For reactions involving gases, pressures are used in place of concentrations, and the expression is based on partial pressures. It is critical to note that only gases and aqueous solutions appear in the expression; solids and liquids, being in their pure phases, do not change their concentrations over the course of the reaction and are therefore omitted.
In the given exercise, the equilibrium constant expression for the dehydration of copper(II) sulfate pentahydrate would only include the water vapor, which is the gaseous product, while the hydrated copper(II) sulfate, being solid, is excluded. Therefore, the expression simplifies to K_e = [H₂O(g)]².

Partial Pressure Calculation

Partial pressure is a term often encountered in gas-related equilibrium problems. It refers to the pressure exerted by a single gas in a mixture of gases. Dalton's Law states that the total pressure of a mixture of gases is the sum of the partial pressures of each individual gas. When calculating the equilibrium constant using partial pressures, we denote it as K_p.
In the context of our exercise, we are aiming to find the partial pressure of water vapor, which at equilibrium, will be squared to match the equilibrium constant, K_e. In essence, to solve for the unknown partial pressure, we take the square root of K_e. After determining the partial pressure in atmospheres (atm), we may need to convert it to a more conventional unit like millimeters of mercury (mmHg) to align with standard pressure measurements used in the lab or real-world situations.

Efflorescence in Chemistry

Efflorescence is a fascinating phenomenon commonly observed with certain hydrate compounds. It occurs when a hydrated salt loses its water molecules to the atmosphere, often resulting in a powdery or crusty appearance on the surface of the substance. This process is particularly notable in climates with low humidity.
In the case of copper(II) sulfate pentahydrate, efflorescence would involve the loss of two water molecules to form the trihydrate. The tendency of a hydrated salt to effloresce is dependent on the humidity of the surrounding air and the vapor pressure of water at the surface of the salt. If the partial pressure of water in the air is lower than the vapor pressure of water in the salt, efflorescence will occur. The exercise prompts us to calculate the maximum pressure water vapor can exert in the atmosphere without causing efflorescence; simplistically, this is the point at which the hydrate and the environment reach a moisture equilibrium. Our understanding of the equilibrium constant expression and partial pressure calculations directly informs our ability to solve problems involving efflorescence, making these concepts not only theoretically significant but also practically applicable.