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 \(\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) (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 kcal, option (d).

Step 1: Understand the relationship between ΔH and ΔE

The relationship between \( \Delta H \) (enthalpy change) and \( \Delta E \) (internal energy change) for a reaction under constant pressure and temperature is given by \( \Delta H = \Delta E + \Delta n_g RT \), where \( \Delta n_g \) is the change in moles of gas.

Step 2: Compute Δn_g

\( \Delta n_g \) is the change in the number of moles of gases during the reaction. For the reaction, products have \(2 + 3 = 5\) moles of gases, and reactants have \(1 + 2 = 3\) moles of gases. Therefore, \( \Delta n_g = 5 - 3 = 2 \).

Step 3: Calculate ΔE using the formula

We use the formula \( \Delta H = \Delta E + \Delta n_g RT \). Given \( \Delta H = 19.0 \) kcal, \( \Delta n_g = 2 \), \( R = 2.0 \ \text{cal} \ \text{K}^{-1} \ \text{mol}^{-1} \), and \( T = 27 + 273 = 300 \ \text{K} \) (convert to Kelvin):\[ \Delta E = \Delta H - \Delta n_g RT = 19.0 \ - 2 \times 2.0 \times 300 \times 10^{-3} \ \text{kcal} \]Convert cal to kcal by multiplying by \(10^{-3}\).

Step 4: Perform the calculation

Substitute the values into the equation:\[ \Delta E = 19.0 - 2 \times 2.0 \times 300 \times 10^{-3} \]Calculate:\[ \Delta E = 19.0 - 1.2 = 17.8 \text{kcal} \]

Step 5: Conclude the correct option

After calculating \( \Delta E \), the closest option from the given list is \( (d) 17.8 \text{kcal} \). This is the correct answer.

Key Concepts

Understanding Enthalpy Change (ΔH)

Enthalpy change (94H) is a crucial concept in thermodynamics. It represents the total heat content of a system at constant pressure. Enthalpy change measures how much heat is absorbed or released during a chemical reaction.
It is symbolized by 94H and typically measured in kilojoules or kilocalories.

Some key points about enthalpy change include:

• When 94H is positive, the reaction absorbs heat from surroundings, known as an endothermic reaction.
• When 94H is negative, the reaction releases heat, classified as an exothermic reaction.

Understanding 94H helps determine if and how much heat energy is transferred in a reaction. This transfer of heat is essential in predicting reaction behavior, especially in complex chemical systems.
In the context of our exercise, 94H was given as 19.0 kcal at 276C, indicating an endothermic reaction where heat is absorbed as the reaction proceeds.

Inside Internal Energy Change (ΔE)

Internal energy change (94E) refers to the difference in a system's total energy after a reaction compared to before. It comes from the energy present in chemical bonds, kinetic energy, and potential energy within the system.
This concept is vital in understanding how energy transitions impact a system beyond mere heat exchange.

The key points to remember about internal energy change include:

• 94E accounts for changes in all energy types within a system, including heat and work done.
• In chemical reactions, 94E provides insights into the energy involved in breaking and forming bonds.

In our exercise, the relationship between 94H and 94E is captured by the equation 94H = 94E + 94n_gRT, where 94n_g reflects the change in moles of gas, and RT incorporates the gas constant (R) and the temperature expressed in Kelvin (T).
For the given reaction, solving for 94E involved understanding how these components interact, leading to a computed 94E of 17.8 kcal.

Exploring the Ideal Gas Law (PV=nRT)

The ideal gas law is a fundamental equation in thermodynamics representing the state of a hypothetical ideal gas. It is expressed as PV=nRT, where P stands for pressure, V for volume, n for the number of moles, R for the gas constant, and T for temperature in Kelvin.
This law connects these variables and allows predictions about how changes in one will affect the others.

Here's what you should know about the ideal gas law:

• It is a simplification and does not apply to real gases under extreme conditions (e.g., very high pressure or low temperature).
• It helps approximate behaviors for many gases under moderate conditions.
• The gas constant R varies based on units used but is given as approximately 8.314 J/(mol3K) or in this exercise, R is defined as 2.0 cal/(mol3K).

The ideal gas law is closely linked to thermodynamic calculations, as seen in our exercise. It was transformed for use in 94H and 94E relationship calculations, showing the usefulness in evaluating reactions that involve gases. By understanding the principles of the ideal gas law, one can uncover important insights into diverse chemical processes and predict how a gas will react to changes in temperature, pressure, or volume.