Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Given : (I) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\); \(\Delta \mathrm{H}_{298 \mathrm{~K}}^{\circ}=-285.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (II) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\); $$ \Delta \mathrm{H}_{298 \mathrm{~K}}^{\circ}=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1} $$ The molar enthalpy of vapourisation of water will be : [Main Online April 9, 2013] (a) \(241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(22.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(44.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(527.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The molar enthalpy of vaporization of water is \(44.1 \mathrm{~kJ/mol}\). Choice (c).

Step 1: Understand the problem

We need to calculate the molar enthalpy of vaporization of water using the given enthalpy changes for the formation of water in liquid and gaseous states.

Step 2: Recall the definition

The molar enthalpy of vaporization (\(\Delta \mathrm{H}_{vap}\)) is the energy required to convert one mole of a substance from a liquid to a gas.

Step 3: Identify the relevant reactions

The reaction (I) represents the formation of liquid water: \(\mathrm{H}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) with \(\Delta \mathrm{H} = -285.9 \mathrm{~kJ/mol}\), and reaction (II) represents the formation of gaseous water: \(\mathrm{H}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}(g)\) with \(\Delta \mathrm{H} = -241.8 \mathrm{~kJ/mol}\).

Step 4: Calculate the enthalpy change for vaporization

The enthalpy change for the conversion of water from liquid to gas (vaporization) can be found by the difference: \(\Delta \mathrm{H}_{vap} = \Delta \mathrm{H}_{f}(g) - \Delta \mathrm{H}_{f}(l)\). Substitute the given values: \(\Delta \mathrm{H}_{vap} = -241.8 - (-285.9) = 44.1 \mathrm{~kJ/mol}\).

Key Concepts

Thermochemistry

Thermochemistry revolves around the study of energy changes that occur during chemical reactions and physical transformations. It's a captivating branch of chemistry that connects the heart of chemical reactions with energy. When a reaction occurs, energy can be released or absorbed, and this energy is commonly termed as heat.

• Exothermic reactions release heat, making the surroundings warmer. The process of forming water from hydrogen and oxygen, as shown in the given reactions, is such an example.
• Endothermic reactions absorb heat, cooling the surroundings.

In thermochemistry, the focus is not only on whether energy is absorbed or released but also on quantifying it. This quantification helps chemists understand reaction dynamics and predict reaction behaviors under different conditions. As we explore the concept further, especially with enthalpy change and vaporization, you'll see how it all ties together.

Enthalpy Change

Enthalpy change (\( \Delta H \)) is crucial in understanding how energy is absorbed or released during a reaction. Enthalpy is a measure of the total energy within a system or a substance. A change in enthalpy indicates that energy has been either emitted or required during the process.
The calculations provided in the exercise illustrate the difference in enthalpy when forming water as a liquid and as a gas:

• For liquid water formation, the enthalpy change is -285.9 kJ/mol.
• For gaseous water, it is -241.8 kJ/mol.

The negative sign signifies exothermic reactions where energy is lost to the surroundings. Understanding these changes helps explain why reactions are favorable or what energy input might be required to transform substances.

Formation Reaction

A formation reaction is a type of chemical reaction where one mole of a substance is formed from its elements in their standard states. In the provided problem, we have a formation reaction for both liquid and gaseous water.
These reactions are fundamental because they allow us to calculate other essential properties such as the enthalpy of vaporization, as demonstrated in the solution.

• Formation of liquid water: \( \mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \)
• Formation of gaseous water: \( \mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) \)

The difference in enthalpy for these reactions gives insight into the energy involved in phase changes, which is an exciting aspect of thermodynamic studies.

Energy Conversion

Energy conversion is a key player in understanding phase changes, like those from liquid to gas. It involves the transition of energy from one form to another. In our exercise, the focus is on the conversion required during the phase change of water, known as the enthalpy of vaporization.
The enthalpy of vaporization is a prime example of energy conversion. It explains how much energy is necessary for one mole of water to transition from a liquid to a gas, requiring the surrounding environment to provide energy.It's calculated using the difference in energy release (enthalpy) for the formation of water in liquid and gaseous states. Here, the calculation \( \Delta H_{vap} = -241.8 \mathrm{~kJ/mol} - (-285.9 \mathrm{~kJ/mol}) = 44.1 \mathrm{~kJ/mol} \) describes this required energy transformation.Understanding these conversions helps us predict and manipulate conditions in industrial processes, environmental systems, and daily activities that include energy shifts or requirements.