Question

Match the column I with column II and mark the appropriate choice. $$ \begin{array}{|l|l|l|l|} \hline \text { (A) } & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(g)} & \text { (i) } & \Delta H=\Delta U-2 R T \\ \hline \text { (B) } & \mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(g)} & \text { (ii) } & \Delta H=\Delta U+3 R T \\ \hline \text { (C) } & \mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)} & \text { (iii) } & \Delta H=\Delta U \\ \hline \text { (D) } & \begin{array}{l} 2 \mathrm{~N}_{2} \mathrm{O}_{5(g)} \rightarrow 4 \mathrm{NO}_{2(g)} \\ +\mathrm{O}_{2(g)} \end{array} & \text { (iv) } & \Delta H=\Delta U+R T \\ \hline \end{array} $$ (a) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (ii), (D) \(\rightarrow\) (iv) (b) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (iv), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (ii) (c) (A) \(\rightarrow\) (ii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (iii) (d) (A) \(\rightarrow\) (iv), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (iii)

Short answer

The correct match is: (A) \( \rightarrow \) (iii), (B) \( \rightarrow \) (iv), (C) \( \rightarrow \) (i), and (D) \( \rightarrow \) (ii), which corresponds to option (d).

Step-by-step solution

- Understanding the Relationship Between \( \Delta H \) and \( \Delta U \)

The change in enthalpy (\( \Delta H \) can be related to the change in internal energy (\( \Delta U \) through the expression \( \Delta H = \Delta U + \Delta n_{gas}RT \) where \( \Delta n_{gas} \) is the change in moles of gas and \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin. For a given chemical reaction, \( \Delta n_{gas} \) is determined by the difference in moles of gaseous products and reactants.

- Analyzing Reaction (A)

For reaction (A), \( \mathrm{H}_{2} + \mathrm{Br}_{2} \rightarrow 2\mathrm{HBr} \), the number of moles of gases on both sides is 2, meaning \( \Delta n_{gas} = 0 \). Thus, \( \Delta H = \Delta U \) for reaction (A). Match (A) with (iii).

- Analyzing Reaction (B)

For reaction (B), \( \mathrm{PCl}_{5} \rightarrow \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \), the number of moles of gas decreases from 1 to 2, so \( \Delta n_{gas} = 1 \). Therefore, \( \Delta H = \Delta U + R T \) for reaction (B). Match (B) with (iv).

- Analyzing Reaction (C)

For reaction (C), \( \mathrm{N}_{2} + 3\mathrm{H}_{2} \rightarrow 2\mathrm{NH}_{3} \), the number of moles of gases decreases from 4 to 2, and \( \Delta n_{ gas} = -2 \). Hence, \( \Delta H = \Delta U - 2R T \) for reaction (C). Match (C) with (i).

- Analyzing Reaction (D)

For reaction (D), \( 2\mathrm{N}_{2}\mathrm{O}_{5} \rightarrow 4\mathrm{NO}_{2} + \mathrm{O}_{2} \), the number of moles of gases remains the same, with 2 moles on both sides. So \( \Delta n_{gas} = 0 \) and \( \Delta H = \Delta U \) for reaction (D). Match (D) with (iii). However, since reaction (A) also matches (iii), and each reaction should match a unique option, ensure to recheck the reactions.

- Correction for Reaction (D)

Re-examine (D) because there must be a mistake in the previous matching. Counting again, there are 2 moles of reactant gas turning into 5 moles of product gas (since there is an additional \( \mathrm{O}_{2} \) produced). The \( \Delta n_{gas} \) for reaction (D) is \( 3 \) thus, \( \Delta H = \Delta U + 3RT \) for reaction (D). Match (D) with (ii).

- Final Matching

With the corrected analysis for reaction (D), the complete matching should be: (A) \( \rightarrow \) (iii), (B) \( \rightarrow \) (iv), (C) \( \rightarrow \) (i), and (D) \( \rightarrow \) (ii). The appropriate choice is (d).

Key concepts

Enthalpy Change

Enthalpy change, symbolized as \( \Delta H \), is one of the fundamental concepts in thermochemistry. It represents the total heat content of a system and is directly related to the heat that is either absorbed or released during a chemical reaction under constant pressure.

For instance, if a chemical reaction emits heat, it is considered exothermic and the \( \Delta H \) would be negative, indicating that the system lost heat to its surroundings. Conversely, reactions that absorb heat are endothermic, leading to a positive \( \Delta H \).

In the context of gases, which are highly affected by changes in temperature and pressure, enthalpy change not only includes internal energy variations but also the work done when the system expands or contracts. The relationship can be expressed mathematically as:
\[ \Delta H = \Delta U + \Delta n_{gas}RT \]
where \( \Delta U \) is the change in internal energy, \( \Delta n_{gas} \) is the change in moles of gas during the reaction, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

Understanding the enthalpy change in chemical reactions enables students to predict whether a substance will feel hot or cold to the touch during a reaction, which is a compelling illustration of energy transfer in the everyday world.

Internal Energy

Internal energy, denoted by \( \Delta U \), accounts for the energy contained within a system. It incorporates the kinetic energy due to the motion of the particles, as well as the potential energy resulting from the attractions and repulsions among particles.

Chemical reactions involve breaking and forming bonds, actions which inherently change the system's internal energy. This change can occur as breaking bonds typically requires energy (endothermic) and forming bonds releases energy (exothermic).

In the realm of thermochemistry, determining the change in internal energy helps us to comprehend energy conservation during reactions. For instance, though energy may be absorbed or released, the total internal energy change signifies the energy that remains within the system or is transferred to the surroundings after the reaction completes.

If we look at our textbook exercise, the change in internal energy plays a critical role, as it helps to determine the enthalpy change when considering the moles of gases involved in the reactions and provided that temperature is consistent throughout the process.

Chemical Reactions

Chemical reactions represent the transformation of substances, involving the rearrangement of atoms and changes in their chemical bonds. These reactions are described using chemical equations that include reactants (starting materials) and products (substances formed).

During a chemical reaction, energy is either released or absorbed, as evidenced by the changes in enthalpy and internal energy. It's essential to grasp that although substances are rearranged, atoms are neither created nor destroyed (law of conservation of mass).

Approaching these reactions step by step, such as in our textbook solution, provides better insight into the complexities of these processes. For instance, balancing the number of atoms for each element, considering the physical states of the reactants and products, and calculating the heat exchange with the surroundings are all fundamental steps in comprehending how chemical reactions proceed and are depicted through thermochemical equations.

Moles of Gases

In chemistry, the mole concept is crucial for quantifying the amount of substance. One mole corresponds to Avogadro's number of particles, which is \( 6.022 \times 10^{23} \) particles. For gases, it is particularly important as it helps link microscopic behavior to macroscopic properties via the ideal gas law.

When dealing with chemical equations, especially those involving gases, it's important to calculate the change in moles of gases (\( \Delta n_{gas} \)). This calculation involves finding the difference in the number of moles of gaseous reactants and products. This number helps derive the enthalpy change when coupled with the other variables such as temperature and the universal gas constant.

As demonstrated in our solved exercise, understanding the concept of \( \Delta n_{gas} \) is vital. For example, a reaction producing more gas molecules than it consumes will generally cause the system to do work on its surroundings, which is considered during the calculation of enthalpy change, as seen in the equation \( \Delta H = \Delta U + \Delta n_{gas}RT \).