Problem 158 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) carried out at constant temperature and pressure. If \(\Delta \mathrm{H}\) and \(\Delta \mathrm{U}\) are the enthalpy and internal energy changes for the reaction, which of the following expressions is true? (a) \(\Delta \mathrm{H}=0\) (b) \(\Delta \mathrm{H}=\Delta \mathrm{U}\) (c) \(\Delta \mathrm{H}<\Delta \mathrm{U}\) (d) \(\Delta \mathrm{H}>\Delta \mathrm{U}\)

Step-by-Step Solution

Answer

Option (c), \(\Delta H < \Delta U\), is correct.

Step 1: Recall the Relationship between ΔH and ΔU

ΔH (enthalpy change) and ΔU (internal energy change) are related by the formula \[ \Delta H = \Delta U + \Delta nRT \]where Δn is the change in moles of gas, R is the ideal gas constant, and T is the temperature (in Kelvin). This relationship holds under constant pressure conditions.

Step 2: Determine Δn for the Reaction

The reaction is \(\mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\). Calculate the change in moles of gas (Δn):- Initial moles of gas: \(1 + 3 = 4\)- Final moles of gas: \(2\)Thus, \(\Delta n = 2 - 4 = -2\).

Step 3: Apply the Relationship Formula

Substitute Δn into the relationship formula:\[ \Delta H = \Delta U + (-2)RT \]This simplifies to:\[ \Delta H = \Delta U - 2RT \]This indicates that \(\Delta H < \Delta U\) because \(-2RT\) is subtracted from \(\Delta U\).

Step 4: Conclusion

From the expression \(\Delta H = \Delta U - 2RT\), we observe that \(\Delta H\) is less than \(\Delta U\) because the term \(-2RT\) is negative. Thus, option (c), \(\Delta H < \Delta U\), is the correct choice.

Key Concepts

Internal Energy Change

The concept of internal energy change, represented as \( \Delta U \), is vital when discussing chemical reactions. Internal energy is the sum total of kinetic and potential energies of all particles in a system. During a chemical reaction, bonds are broken and formed, leading to changes in the internal energy of the substances involved.

In the given reaction, \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \), the internal energy change \( \Delta U \) is an important factor to consider. The relationship between internal energy change and enthalpy change (\( \Delta H \)) is given by the formula: \[ \Delta H = \Delta U + \Delta nRT \]Under constant pressure and temperature, this formula helps us understand how much energy is absorbed or released as heat during the reaction. By knowing \( \Delta U \), you can calculate \( \Delta H \), and understand the energy dynamics of the reaction.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics, expressed as \( PV = nRT \). It describes how gases behave under various conditions of pressure (\( P \)), volume (\( V \)), and temperature (\( T \)), with \( n \) representing moles of gas, and \( R \) as the ideal gas constant.

In the context of this reaction, understanding the Ideal Gas Law aids in examining the role gases play in reaction dynamics. Particularly, it helps us incorporate the \( \Delta nRT \) term in the relationship between \( \Delta H \) and \( \Delta U \). Here, \( \Delta n \) signifies the change in the number of moles of gas, which, when paired with the temperature and gas constant, influences the enthalpy of the reaction. Realizing how the Ideal Gas Law integrates these variables helps us predict how changes in any of these quantities can impact the reaction's energy profile.

Change in Moles of Gas

The change in moles of gas, represented as \( \Delta n \), is crucial when assessing chemical reactions involving gases. It determines the shift in the total number of gaseous particles before and after the reaction.

In the reaction given, \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \), calculating \( \Delta n \) involves comparing the moles of reactants and products. Initially, there are 1 mole of \( \mathrm{N}_2 \) and 3 moles of \( \mathrm{H}_2 \), totaling 4 moles of gas. Post-reaction, there are 2 moles of \( \mathrm{NH}_3 \). Therefore, \( \Delta n = 2 - 4 = -2 \). Simple subtraction gives the change in gas moles.

This change \( \Delta n \) is plugged into the equation \( \Delta H = \Delta U + \Delta nRT \), helping determine if the reaction releases or absorbs heat. Knowing that \( \Delta n = -2 \), it implies a net reduction in the number of moles as the reaction progresses, and this reduction influences the enthalpy change by modifying the \( RT \) term subtracted from \( \Delta U \).