The following reactions carried in open vessel. The reaction for which `Delta H = Delta U` will be

Options

A. `PCl_(5)(g) rightarrow PCl_(3)(g) + Cl_(2)(g)`
B. `2CO(g) + O_(2)(g) rightarrow 2CO_(2)(g)`
C. `N_(2)(g) + 3H_(2)(g) rightarrow 2NH_(3)(g)`
D. `H_(2)(g) + I_(2)(g) rightarrow 2HI(g)`

Answer

To determine which reaction has ΔH = ΔU, we need to analyze the change in the number of moles of gaseous products and reactants (ΔNg) for each reaction. The relationship between ΔH and ΔU is given by the equation:

\[ \Delta H = \Delta U + \Delta N_g RT \]

Where:
- ΔH = change in enthalpy
- ΔU = change in internal energy
- ΔNg = N2 - N1 (the difference between the number of moles of gaseous products and gaseous reactants)
- R = universal gas constant
- T = temperature in Kelvin

For ΔH to equal ΔU, ΔNg must be zero (ΔNg = 0).

Now, let’s analyze each reaction step by step:

### Step 1: Analyze the First Reaction
**Reaction:** PCl5(g) ⇌ PCl3(g) + Cl2(g)

- Products: 1 mole of PCl3 + 1 mole of Cl2 = 2 moles
- Reactants: 1 mole of PCl5 = 1 mole

\[ \Delta N_g = N_{products} - N_{reactants} = 2 - 1 = 1 \]

**Conclusion:** ΔNg ≠ 0, so ΔH ≠ ΔU.

### Step 2: Analyze the Second Reaction
**Reaction:** 2 CO(g) + O2(g) → 2 CO2(g)

- Products: 2 moles of CO2 = 2 moles
- Reactants: 2 moles of CO + 1 mole of O2 = 3 moles

\[ \Delta N_g = 2 - 3 = -1 \]

**Conclusion:** ΔNg ≠ 0, so ΔH ≠ ΔU.

### Step 3: Analyze the Third Reaction
**Reaction:** N2(g) + 3 H2(g) → 2 NH3(g)

- Products: 2 moles of NH3 = 2 moles
- Reactants: 1 mole of N2 + 3 moles of H2 = 4 moles

\[ \Delta N_g = 2 - 4 = -2 \]

**Conclusion:** ΔNg ≠ 0, so ΔH ≠ ΔU.

### Step 4: Analyze the Fourth Reaction
**Reaction:** H2(g) + I2(g) ⇌ 2 HI(g)

- Products: 2 moles of HI = 2 moles
- Reactants: 1 mole of H2 + 1 mole of I2 = 2 moles

\[ \Delta N_g = 2 - 2 = 0 \]

**Conclusion:** ΔNg = 0, so ΔH = ΔU.

### Final Answer:
The reaction for which ΔH = ΔU is the fourth reaction: **H2(g) + I2(g) ⇌ 2 HI(g)**.

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