Problem 78 - A Complete Resource Book in Chemistry for JEE Main

Question

The enthalpy change \((\Delta \mathrm{H})\) for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) is \(-92.38 \mathrm{~kJ}\) at 298 \(\mathrm{K}\). The internal energy change \(\Delta \mathrm{U}\) at \(298 \mathrm{~K}\) is (a) \(-92.38 \mathrm{~kJ}\) (b) \(-87.42 \mathrm{~kJ}\) (c) \(-97.34 \mathrm{~kJ}\) (d) \(-89.9 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The internal energy change () is -97.34  (option c).

Step 1: Understand the Relationship

The relationship between enthalpy change ( ) and internal energy change () is given by the equation:  =  + RT, where  is the change in the number of moles of gas,  is the ideal gas constant (.314 //), and  is the temperature in Kelvin.

Step 2: Calculate Change in Moles

Determine , which is the change in the number of moles of gas during the reaction. In the equation:  + 3 -> 2, the total moles of gas decrease from 4 to 2. Thus,  = 2 - 4 = -2.

Step 3: Apply the Formula

Substitute the values into the formula:  =  + RT. Here,  = -92.38 ,  = -2,  = 8.314 //, and  = 298 . Calculate RT, which equals to -2  8.314  298  = -4953.2 , or -4.953  (since 1 / /1000 ).

Step 4: Solve for Internal Energy Change

Now substitute the calculated RT value into the equation:  = -92.38  + (-4.953 ) = -92.38  + 4.953  = -97.333 .

Key Concepts

Enthalpy Change

When you hear the term **enthalpy change**, think of it as the heat transferred in a reaction at constant pressure.This is represented by the symbol \(\Delta H\) and measured in kilojoules (kJ).The enthalpy change provides a clear understanding of whether a reaction is exothermic or endothermic.
- **Exothermic Reaction:** The system releases heat, and \(\Delta H\) is negative. For example, the formation of ammonia (NH₃) from nitrogen and hydrogen gases releases energy, which is evident from its enthalpy change of \(-92.38 \space \text{kJ}\).- **Endothermic Reaction:** In this case, the system absorbs heat and \(\Delta H\) is positive.
Understanding enthalpy change is vital for predicting the energy impacts of chemical reactions, which is a key factor in many industrial applications.

Internal Energy Change

The **internal energy change**, denoted as \(\Delta U\), depicts the total energy change within a system.It considers all forms of energy including kinetic and potential energy. The change in internal energy relates closely to enthalpy change, especially when pressure and volume transformations occur.
The connection between enthalpy (\(\Delta H\)) and internal energy is given by:\[\Delta U = \Delta H - \Delta nRT\]Here:

• \(\Delta n\) stands for the change in moles of gas.
• \(R\) represents the ideal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.

In the ammonia formation reaction, internal energy change was calculated as \(-97.34 \text{kJ}\) using the relationship between \(\Delta H\) and the expression for work done related to gas compression or expansion terms.

Haber Process

The **Haber Process** is an industrial marvel that produces ammonia by reacting nitrogen and hydrogen gases under high pressure and specific temperatures. Developed by Fritz Haber and Carl Bosch, this method allowed for mass production of ammonia, which is crucial for fertilizers and industrial processes.
Here's why the Haber Process is significant:

• **Critical for Agriculture:** Ammonia is a vital component in fertilizers, aiding in crop growth and agricultural efficiency.
• **Energy Involvement:** Understanding enthalpy and internal energy changes helps in optimizing the conditions necessary for maximum ammonia yield.
• **Environmental Impact:** Though ammonia production is essential, its industrial process requires substantial energy, often sourced from fossil fuels, highlighting the need for energy-efficient methods.

The process balances yield and energy consumption by carefully controlling the reaction conditions, using catalysts and maintaining suitable pressure and temperature levels.