Problem 110 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+3 \mathrm{D}(\mathrm{g})\) The value of \(\Delta \mathrm{H}\) at \(27^{\circ} \mathrm{C}\) is \(19.0 \mathrm{kcal}\). The value of \(\Delta \mathrm{E}\) for the reaction would be (given \(\left.\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(20.8 \mathrm{kcal}\) (b) \(19.8 \mathrm{kcal}\) (c) \(18.8 \mathrm{kcal}\) (d) \(17.8 \mathrm{kcal}\)

Step-by-Step Solution

Answer

The value of \(\Delta E\) for the reaction is 17.8 \mathrm{kcal} (option d).

Step 1: Understanding the Given Information

We need to determine the change in internal energy, \(\Delta E\), for the given reaction at \(27^{\circ} \mathrm{C}\). We are given \(\Delta H = 19.0 \mathrm{kcal}\) and the universal gas constant \(R = 2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\). The reaction is \(\mathrm{A} + 2\mathrm{B} \rightarrow 2\mathrm{C} + 3\mathrm{D}\).

Step 2: Identify the Relation Between ΔH and ΔE

The relation is given by: \( \Delta H = \Delta E + \Delta n_gRT \), where \(\Delta n_g\) is the change in moles of gas, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

Step 3: Calculate Δn_g

Find \(\Delta n_g\) (change in moles of gas). The moles of gaseous reactants \(= 1 + 2 = 3\) and the moles of gaseous products \(= 2 + 3 = 5\). Thus, \(\Delta n_g = 5 - 3 = 2\).

Step 4: Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin: \(T = 27 + 273 = 300\, \mathrm{K}\).

Step 5: Compute Δn_gRT

Substitute \(\Delta n_g = 2\), \(R = 2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\) (converted to \(0.002 \mathrm{kcal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\)), and \(T = 300 \mathrm{K}\) into \(\Delta n_gRT\):\[\Delta n_gRT = 2 \times 0.002 \times 300 = 1.2 \mathrm{kcal}\].

Step 6: Calculate ΔE

Use the equation \( \Delta H = \Delta E + \Delta n_gRT \) to find \(\Delta E\):\[19.0 = \Delta E + 1.2\], solving for \(\Delta E\) gives:\[\Delta E = 19.0 - 1.2 = 17.8 \mathrm{kcal}\].

Step 7: Answer the Multiple Choice

Based on the calculation of \(\Delta E\), the correct answer is (d) \(17.8 \mathrm{kcal}\).

Key Concepts

Enthalpy Change (ΔH)

In thermodynamics, enthalpy change, represented by \( \Delta H \), is a crucial concept when examining chemical reactions. Enthalpy is a measure of the total energy of a thermodynamic system, which includes internal energy and pressure-volume work. When a reaction occurs at constant pressure, the heat absorbed or released is equal to the change in enthalpy.

In the presented exercise, \( \Delta H \) is given as \( 19.0 \text{kcal} \). This indicates that the reaction absorbs heat under constant pressure, meaning it is endothermic. Understanding \( \Delta H \) helps us determine whether a reaction is endothermic (absorbs heat) or exothermic (releases heat).

• Endothermic Reaction: Absorbs heat, \( \Delta H > 0 \).
• Exothermic Reaction: Releases heat, \( \Delta H < 0 \).

Understanding the concept of enthalpy change provides insight into the energy flow in a reaction, allowing us to predict how the reaction will behave under various conditions.

Internal Energy Change (ΔE)

Internal energy change, depicted by \( \Delta E \), represents the change in the total internal energy of a system during a reaction. It includes all forms of energy contained in the system, such as kinetic and potential energy at the atomic level.

The relationship between \( \Delta H \) and \( \Delta E \) is given by the equation:
\[ \Delta H = \Delta E + \Delta n_gRT \]
where \( \Delta n_g \) is the change in moles of gas, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. This relationship is derived from the consideration that enthalpy accounts for work done against atmospheric pressure, while internal energy does not.

In the example provided, we calculated \( \Delta E \) by rearranging the equation to:
\[ \Delta E = \Delta H - \Delta n_gRT \]
By substituting known values, we found \( \Delta E \) to be \( 17.8 \text{kcal} \). This calculation allows us to understand how changes in moles of gas and temperature impact the internal energy of the system.

Gas Laws

Gas laws play a significant role in understanding how gases behave and react, especially in thermochemical processes. In the given exercise, the ideal gas law is implicitly employed when dealing with moles of gases and temperature.

The ideal gas equation is given by:
\[ PV = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. This equation helps determine the state of a certain amount of gas under specific conditions.

• Pressure: Influences the arrangement of gas molecules within a volume.
• Volume: The space that gas molecules occupy.
• Temperature: A measure of the kinetic energy of gas molecules.

In the context of the exercise, understanding gas laws helps us calculate \( \Delta n_gRT \), the change in moles of gas times the gas constant and temperature, which is necessary to determine \( \Delta E \). By applying the principles of gas behavior, we can accurately determine changes in energy associated with the reaction.