Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

For which of the following reactions, is \(\Delta H\) equal to \(\Delta E ?\) (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) (d) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\)

Step-by-Step Solution

Answer

For reactions (a) and (d), \(\Delta H = \Delta E\).

Step 1: Understanding Enthalpy and Internal Energy

The change in enthalpy (\(\Delta H\)) is related to the change in internal energy (\(\Delta E\)) by the equation \(\Delta H = \Delta E + \Delta (PV)\), where \(P\) is pressure and \(V\) is volume. For gases, \(\Delta (PV)\) can be expressed as \(\Delta nRT\), where \(\Delta n\) is the change in the number of moles of gas, \(R\) is the gas constant, and \(T\) is the temperature. When \(\Delta n = 0\), \(\Delta H = \Delta E\).

Step 2: Calculate change in moles for reaction (a)

Reaction (a) is \(\mathrm{H}_{2}( ext{g}) + \mathrm{I}_{2}( ext{g}) \rightarrow 2 \mathrm{HI}( ext{g})\). The initial moles are 1 (\(\mathrm{H}_2\)) + 1 (\(\mathrm{I}_2\)) = 2 and final moles are 2 (\(\mathrm{HI}\)). Therefore, \(\Delta n = 2 - 2 = 0\).

Step 3: Calculate change in moles for reaction (b)

Reaction (b) is \(\mathrm{PCl}_{5}( ext{g}) \rightarrow \mathrm{PCl}_{3}( ext{g}) + \mathrm{Cl}_{2}( ext{g})\). The initial moles are 1 (\(\mathrm{PCl}_5\)) and final moles are 1 (\(\mathrm{PCl}_3\)) + 1 (\(\mathrm{Cl}_2\)) = 2. Therefore, \(\Delta n = 2 - 1 = 1\).

Step 4: Calculate change in moles for reaction (c)

Reaction (c) is \(2 \mathrm{H}_{2} \mathrm{O}_{2}( ext{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}( ext{g}) + \mathrm{O}_{2}( ext{g})\). Initial moles are 2 (\(\mathrm{H}_2\mathrm{O}_2\)) and final moles are 2 (\(\mathrm{H}_2\mathrm{O}\)) + 1 (\(\mathrm{O}_2\)) = 3. Therefore, \(\Delta n = 3 - 2 = 1\).

Step 5: Calculate change in moles for reaction (d)

Reaction (d) is \(\mathrm{C}( ext{s}) + \mathrm{O}_{2}( ext{g}) \rightarrow \mathrm{CO}_{2}( ext{g})\). The initial moles of gas are 1 (\(\mathrm{O}_2\)) and final moles are 1 (\(\mathrm{CO}_2\)). Therefore, \(\Delta n = 1 - 1 = 0\).

Step 6: Identify reactions where \(\Delta n = 0\)

For reactions where \(\Delta n = 0\), \(\Delta H = \Delta E\). From the calculations: reaction (a) \(\Delta n = 0\), reaction (b) \(\Delta n = 1\), reaction (c) \(\Delta n = 1\), and reaction (d) \(\Delta n = 0\). Thus, for reactions (a) and (d), \(\Delta H = \Delta E\).

Key Concepts

Thermochemistry

Thermochemistry is a branch of chemistry focused on the study of energy and heat transformations in chemical reactions. In particular, it explores how energy is transferred in reactions and how this relates to the principles of energy conservation. One of the most important concepts in thermochemistry is enthalpy (H). Enthalpy is a measure of the total energy of a thermodynamic system, including internal energy plus the product of pressure and volume \( (PV) \). It is especially useful when dealing with constant-pressure environments, like many chemical reactions.
To understand the energy changes in a reaction, we often look at the change in enthalpy, H, which tells us whether energy is absorbed or released by the system. A positive H indicates that the system absorbs heat (endothermic reaction), while a negative H suggests that the system releases heat (exothermic reaction). These observations help chemists understand reaction spontaneity, a key aspect of thermodynamics. By studying thermochemistry, we gain insight into the energy profiles of reactions, which is crucial for predicting reaction behavior and efficiency.

Gaseous Reactions

Gaseous reactions involve substances in the gaseous state, where the behavior of the involved molecules can be described by gas laws. The state of gases makes these reactions particularly interesting since they often involve significant volume changes, which are important for understanding reactions at a microscopic level.
In gaseous reactions, the volume change can be linked to changes in pressure and temperature, as well as the initial and final number of moles of gas being involved. The formula (PV) = (nRT) is crucial, where (n) represents the change in moles of gas, \( R \) is the gas constant, and \( T \) is the temperature. A reaction's enthalpy change (H) relates to its internal energy change (E) through exactly this formula.
If you can identify that there is no change in the number of moles across a reaction, you will find that (PV) = 0, leading to scenarios where H = E. Determining (n) is therefore a critical analytical step when examining gaseous reactions. Understanding how molecules interact in the gaseous state allows chemists to predict outcomes and optimize conditions for desired chemical transformations.

Mole Calculation

Mole calculation is a fundamental aspect of chemistry that provides a bridge between the atomic scale and the macroscopic world. A "mole" is a unit that allows chemists to count particles, like atoms or molecules, using a large, manageable number: Avogadro's number, \( 6.022 \times 10^{23} \) entities per mole.
When dealing with chemical equations, calculating moles helps determine how much of each substance participates in the reactions. This is essential for balancing chemical equations and quantifying reactants and products. During reactions, it's important to calculate the change in moles, (n), especially for gaseous reactions, to understand volume changes and to determine whether H equals E.
Performing mole calculations involves using known standards to solve problems regarding mass, volume, concentration, and particle number. By mastering this concept, students can accurately predict chemical quantities and their transformations in reactions, providing insights into chemical processes that are both theoretical and practical.