3. For the reaction, [tex]$PCl_5(g) \longrightarrow PCl_3(g) + Cl_2(g)$[/tex]:

A. [tex]$\Delta H = \Delta U$[/tex]
B. [tex][tex]$\Delta H \ \textgreater \ \Delta U$[/tex][/tex]
C. [tex]$\Delta H \ \textless \ \Delta U$[/tex]
D. None of these



Answer :

Sure, let's analyze the reaction and the relationship between enthalpy (ΔH) and internal energy (ΔU) changes.

The reaction given is:
[tex]\[ \text{PCl}_5(g) \longrightarrow \text{PCl}_3(g) + \text{Cl}_2(g) \][/tex]

Here, PCl₅ decomposes into PCl₃ and Cl₂, both of which are gases.

### Relationship Between ΔH and ΔU
For an ideal gas, the relationship between the enthalpy change (ΔH) and the internal energy change (ΔU) is given by the following equation:
[tex]\[ \Delta H = \Delta U + \Delta nRT \][/tex]

Where:
- Δn is the change in the number of moles of gas.
- R is the universal gas constant.
- T is the absolute temperature.

### Determining Δn:
Δn represents the change in the number of moles of gas during the reaction.
- Number of moles of gas in reactants = 1 (PCl₅)
- Number of moles of gas in products = 1 (PCl₃) + 1 (Cl₂)

Thus,
[tex]\[ \Delta n = (\text{number of moles of gases in products}) - (\text{number of moles of gases in reactants}) \][/tex]
[tex]\[ \Delta n = (1 + 1) - 1 \][/tex]
[tex]\[ \Delta n = 2 - 1 \][/tex]
[tex]\[ \Delta n = 1 \][/tex]

### Analyzing ΔH and ΔU:
Since Δn is positive (Δn = 1):
- The term [tex]\( \Delta nRT \)[/tex] will be positive because both R (gas constant) and T (temperature in Kelvin) are always positive.

Thus, from the equation [tex]\[ \Delta H = \Delta U + \Delta nRT \][/tex], if [tex]\( \Delta nRT \)[/tex] is positive, then:
[tex]\[ \Delta H > \Delta U \][/tex]

Therefore, the correct answer is:

(B) [tex]\( \Delta H > \Delta U \)[/tex]