Question

\(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) ; K_{p}=4 \times 10^{-4} \mathrm{~atm}^{2}\). If the vapour pressure of water is 38 torr then percentage of relative humidity is : (Assume all data at constant temperature) (a) 4 (b) 10 (c) 40 (d) None of these

Short answer

The relative humidity is 40%.

Step-by-step solution

Convert the vapour pressure of water from torr to atm

Since the given equilibrium constant, Kp, is in units of atm, convert the vapour pressure from torr to atm. Use the conversion factor: 1 atm = 760 torr. Divide the vapour pressure of water by 760 torr/atm.

Calculate the partial pressure of water

The vapour pressure of water represents the partial pressure of water at equilibrium. Use the value obtained from Step 1 to represent the partial pressure of H2O(g).

Calculate the relative humidity

Relative humidity is the ratio of the partial pressure of water in air to the equilibrium vapour pressure of water at the same temperature. Express this ratio as a percentage. Since the equilibrium vapour pressure of water is given by Kp and represents the maximum possible pressure at the given temperature, divide the partial pressure of water obtained in Step 2 by the square root of Kp (because Kp represents the square of the pressure units) and multiply by 100 to get the percentage.

Express the relative humidity as a percentage

The relative humidity percentage is the ratio calculated in Step 3 multiplied by 100. This yields the final answer.

Key concepts

Vapour Pressure Conversion

Understanding vapour pressure conversion is essential for solving problems involving gas equilibria and humidity calculations. Vapour pressure is defined as the pressure exerted by a vapour in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. When dealing with various units of pressure like torr and atmospheres (atm), it's crucial to convert them appropriately to ensure consistency in calculations.

For example, atmospheric pressure is a standard measurement unit and is equivalent to 760 torr. To convert from torr to atm, you divide the vapour pressure value by 760. This step is fundamental as it aligns the pressure units with those of the equilibrium constant provided, allowing for accurate calculations in subsequent steps.

Equilibrium Constant (Kp)

The equilibrium constant (Kp) plays a vital role in chemical equilibrium, especially when dealing with gases. It quantifies the ratio of the products to reactants at equilibrium for a reaction involving gases, with each component raised to the power of its stoichiometric coefficient. The 'p' in Kp signifies that the equilibrium concentrations are expressed in terms of partial pressures, typically in atmospheres.

In the context of our exercise, Kp is given for the dehydration reaction of hydrated copper sulfate to form a partially hydrated compound and water vapour. Understanding Kp is crucial because it allows us to connect the known vapour pressure of water with the reaction's dynamics, aiding in the determination of the relative humidity.

Partial Pressure of Water

Partial pressure of water refers to the pressure water vapour contributes to the total pressure in a mixture of gases. According to Dalton's Law of Partial Pressures, in a mixture of non-reacting gases, the total pressure exerted is the sum of the partial pressures of individual gases. In our case, the vapour pressure of water given in the problem statement can be taken directly as the partial pressure of water since it is the only gas present.

Once converted to the appropriate units, this partial pressure serves as a critical component in the formula to calculate relative humidity. It reflects how much water vapour is present compared to how much could potentially exist at the same temperature and pressure, which is what we need to proceed with the next step of calculating the relative humidity.

Chemical Equilibrium

Understand the concept of chemical equilibrium is essential for working with reactions like the one presented in the exercise. At chemical equilibrium, a reaction proceeds at the same rate in both the forward and reverse directions, leading to constant concentrations of reactants and products, though they need not be equal.

In the dehydration reaction of copper sulfate pentahydrate to trihydrate, equilibrium occurs when the rate of water loss to form the trihydrate balances with the rate it is gained back from the vapour phase. Kp is integral here because it indicates the extent to which a reaction proceeds before reaching equilibrium. Its value can also be interpreted to understand the relative amounts of reactants and products present at equilibrium, which in the provided exercise, relates to the presence of water vapour in relation to the solid hydrates of copper sulfate.