Question

The change in entropy, in the conversion of one mole of water at \(373 \mathrm{~K}\) to vapour at the same temperature is (Latent heat of vaporization of water \(=2.257 \mathrm{~kJ} \mathrm{~g}^{-1}\) ) (a) \(99 \mathrm{JK}^{-1}\) (b) \(129 \mathrm{JK}^{-1}\) (c) \(89 \mathrm{JK}^{-1}\) (d) \(109 \mathrm{JK}^{-1}\)

Short answer

The change in entropy is approximately \(109 \mathrm{JK}^{-1}\).

Step-by-step solution

Convert Latent Heat to Joules

Given that the latent heat of vaporization of water is \(2.257 \mathrm{~kJ} \mathrm{~g}^{-1}\). Convert this value from kilojoules per gram to joules per mole. First, convert kilojoules to joules: \(2.257 \mathrm{~kJ} = 2257 \mathrm{~J}\). Since the molar mass of water is 18 g/mol, the latent heat becomes \(2257 \mathrm{~J} \times 18 \mathrm{~g/mol} = 40626 \mathrm{~J/mol}\).

Use the Formula for Entropy Change

The entropy change \( \Delta S \) for a phase change can be calculated using the formula: \[ \Delta S = \frac{\Delta H}{T} \] where \( \Delta H \) is the enthalpy change and \( T \) is the temperature in Kelvin. Substitute the known values \( \Delta H = 40626 \mathrm{~J/mol}\) and \( T = 373 \mathrm{~K}\) into the formula: \[ \Delta S = \frac{40626 \mathrm{~J/mol}}{373 \mathrm{~K}} \].

Calculate the Entropy Change

Perform the division: \[ \Delta S = \frac{40626}{373} \approx 108.96 \mathrm{~J/K} \].

Choose the Closest Answer

Compare the calculated entropy change with the provided options. The calculated value \(108.96 \mathrm{~J/K}\) is closest to option (d) \(109 \mathrm{JK}^{-1}\). Thus, the answer is (d).

Key concepts

Latent Heat of Vaporization

Latent heat of vaporization is the amount of energy required to change a liquid into a vapor without a change in temperature. In this process, the liquid absorbs energy, which is necessary for the molecules to overcome intermolecular forces and escape into the vapor phase. This energy input is crucial for the phase change to occur, and it varies from one substance to another.

For water, the latent heat of vaporization is given as 2.257 kJ/g, meaning that it takes 2.257 kilojoules of energy to vaporize one gram of water. Converting this to joules simplifies calculations, resulting in 2257 J/g. To find out how much energy is needed for one mole, multiply by the molar mass of water (18 g/mol), getting 40626 J/mol. Understanding this conversion is important when dealing with thermodynamic calculations related to phase changes.

Entropy Formula

Entropy is a measure of disorder or randomness in a system. When calculating the entropy change during a phase transition, the formula used is \[ \Delta S = \frac{\Delta H}{T} \]where \( \Delta S \) is the change in entropy, \( \Delta H \) is the enthalpy change, and \( T \) is the absolute temperature in Kelvin. This formula provides insight into how heat exchange influences disorder in a system.

During vaporization, the entropy increases as well. Molecules in the vapor state have more freedom of movement and thus more disorder than those in the liquid state. Plugging known values, like the heat of vaporization (40626 J/mol) and temperature (373 K), into the formula allows you to solve for the change in entropy, crucial for predicting system behavior.

Phase Change

Phase changes occur when a substance changes from one state of matter to another, such as solid to liquid, liquid to gas, or vice versa. During a phase change, energy is transferred in the form of latent heat, causing a transformation without a change in temperature. This energy is used to overcome forces holding molecules together.

In the case of vaporization, molecules in the liquid phase absorb heat, without changing temperature initially, until they have enough energy to break free into the gas phase. As the phase change occurs at constant temperature, this scenario often involves intricate calculations regarding energy and entropy changes. Grasping these transformations helps to appreciate why certain processes, like boiling or freezing, occur as they do.

Mole Calculation

Converting between mass units and moles is a vital skill in chemistry. It's often necessary when dealing with energy calculations involving phase changes. One mole of a substance contains Avogadro's number of molecules, approximately \(6.022 \times 10^{23}\) molecules.

For water, with a molar mass of 18 g/mol, one mole weighs 18 grams. This relationship allows for the conversion of latent heat values from a per gram basis to a per mole basis, simplifying thermodynamic calculations. Such conversions frequently appear in problems dealing with phase changes, energy calculations, and reaction stoichiometry. Fully understanding mole calculations provides a strong foundation for tackling various chemistry and physics problems.