Problem 47 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}\) (I) at constant temperature, \(\Delta \mathrm{H}-\Delta \mathrm{E}\) is (a) \(+3 \mathrm{RT}\) (b) \(-\mathrm{RT}\) (c) \(+\mathrm{RT}\) (d) \(-3 \mathrm{RT}\)

Step-by-Step Solution

Answer

The answer is (c) \(+\mathrm{RT}\).

Step 1: Write Down the Equation

We are given the chemical equation for the combustion of propane: \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})\). We need to determine \(\Delta \mathrm{H} - \Delta \mathrm{E}\).

Step 2: Recall the Relationship Between ΔH and ΔE

We know that \(\Delta \mathrm{H} = \Delta \mathrm{E} + \Delta n_{\text{g}} \cdot \mathrm{RT}\), where \(\Delta n_{\text{g}}\) is the change in moles of gas, and \(\mathrm{R}\) is the ideal gas constant. Our task is to calculate \(\Delta n_{\text{g}}\) and then find \(\Delta \mathrm{H} - \Delta \mathrm{E}\).

Step 3: Calculate Δn_g

\(\Delta n_{\text{g}}\) is calculated as the difference between the moles of gaseous products and the moles of gaseous reactants. For this reaction, gaseous reactants are \(1+5=6\) moles, and gaseous products are \(3+4=7\) moles. Therefore, \(\Delta n_{\text{g}} = 7 - 6 = 1\).

Step 4: Determine ΔH - ΔE

Substitute \(\Delta n_{\text{g}} = 1\) into the equation: \(\Delta \mathrm{H} = \Delta \mathrm{E} + 1 \cdot \mathrm{RT}\). Hence, \(\Delta \mathrm{H} - \Delta \mathrm{E} = \mathrm{RT}\).

Step 5: Choose the Correct Option

From the previous step, we found \(\Delta \mathrm{H} - \Delta \mathrm{E}\) equals \(\mathrm{RT}\). The correct option is (c) \(+\mathrm{RT}\).

Key Concepts

Gibbs Free Energy

Gibbs free energy, often symbolized as \( \Delta G \), is crucial in determining the spontaneity of a chemical reaction. It combines enthalpy and entropy into a single value that predicts whether a reaction will proceed without external input. The fundamental equation for Gibbs free energy is:

• \( \Delta G = \Delta H - T \Delta S \)

where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy. For a reaction to be spontaneous under constant pressure and temperature, \( \Delta G \) must be negative.
This equation tells us how energy and disorder balance out. Enthalpy change relates to heat absorbed or released, while entropy change reflects how system disorder changes during a reaction. Lower Gibbs free energy in the products than in reactants means the reaction tends to occur on its own.

Enthalpy Change

Enthalpy change, denoted by \( \Delta H \), measures the heat change in a system at constant pressure. It is a state function, meaning it depends only on the initial and final states, not on the path taken. This characteristic makes it useful for calculating heat changes in reactions.
The equation \( \Delta H = \Delta E + P \Delta V \) is used, where \( \Delta E \) is internal energy change, \( P \) is pressure, and \( \Delta V \) is change in volume. However, in the context of gaseous reactions where volume can change noticeably, it often simplifies to \( \Delta H = \Delta E + \Delta n_g \cdot RT \). This form is widely exploited in problems involving gaseous reactions.
Exothermic reactions, which release heat, have a negative \( \Delta H \). Conversely, endothermic reactions, which absorb heat, have a positive \( \Delta H \). In the context of understanding chemical energetics, grasping enthalpy change helps predict whether a reaction will release or absorb energy.

Internal Energy

Internal energy, symbolized as \( \Delta E \), represents the total energy contained within a system. It accounts for the sum of kinetic and potential energies of all particles. It's a cornerstone concept in thermodynamics, essential for analyzing energy changes and exchanges.Although internal energy includes various forms of energy, when it comes to chemical reactions, we're usually interested in how it changes due to heat and work interactions. The First Law of Thermodynamics formalizes this idea with:

• \( \Delta E = q + w \)

where \( q \) is the heat exchanged and \( w \) is the work done by or on the system. In reactions involving gases, chemical work often relates to volume changes, thus \( w = -P \Delta V \).Understanding internal energy change is crucial for interpreting reaction contexts, especially when calculating energy flow directions during transformations. It underscores the interconnectedness of heat and work in energy transfer scenarios.