Consider the reaction at 300 K
`H_(2)(g)+Cl_(2)(g)to2HCl(g)" "DeltaH^(@)=-185kJ`
If 3 mole of `H_(2)` completely react with 3 mole of `Cl_(2)` to form HCl. What is `DeltaU^(@)` for this reaction ?

Options

A. 0
B. `-185kJ`
C. 555 kJ
D. None of these

Answer

To solve the problem, we need to calculate the change in internal energy (ΔU) for the reaction given that 3 moles of H₂ react with 3 moles of Cl₂ to form HCl. The reaction is:

\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \]

Given:
- ΔH° = -185 kJ for the reaction as written (which produces 2 moles of HCl from 1 mole of H₂ and 1 mole of Cl₂).

### Step 1: Determine the change in the number of moles of gas (Δn_g)

For the reaction:
- Reactants: 1 mole of H₂ + 1 mole of Cl₂ = 2 moles of gas
- Products: 2 moles of HCl = 2 moles of gas

Thus, the change in the number of moles of gas (Δn_g) is:
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

### Step 2: Calculate the total ΔH for the reaction with 3 moles of H₂ and 3 moles of Cl₂

Since the stoichiometry of the reaction shows that 1 mole of H₂ reacts with 1 mole of Cl₂ to produce 2 moles of HCl, if we have 3 moles of H₂ and 3 moles of Cl₂, the reaction will produce:
\[ 3 \text{ moles of H}_2 \rightarrow 3 \text{ moles of Cl}_2 \rightarrow 6 \text{ moles of HCl} \]

The ΔH for the reaction producing 6 moles of HCl can be calculated as:
\[ \Delta H = -185 \, \text{kJ} \times 3 = -555 \, \text{kJ} \]

### Step 3: Use the relationship between ΔH and ΔU

The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) is given by:
\[ \Delta H = \Delta U + \Delta n_g RT \]

Where:
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = 300 K

Since we found that Δn_g = 0, the equation simplifies to:
\[ \Delta H = \Delta U + 0 \]
Thus:
\[ \Delta U = \Delta H \]

### Step 4: Calculate ΔU

From our calculation:
\[ \Delta U = -555 \, \text{kJ} \]

### Final Answer:
The change in internal energy (ΔU) for the reaction when 3 moles of H₂ react with 3 moles of Cl₂ is:
\[ \Delta U = -555 \, \text{kJ} \]

---