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

Options

A. 0
B. `-185 kJ`
C. -370 kJ
D. None of these

Explanation

To solve the problem, we need to calculate the change in internal energy (ΔU) for the reaction given the change in enthalpy (ΔH) and the change in the number of moles of gas (ΔNG). Here’s a step-by-step solution:

### Step 1: Write the balanced chemical reaction
The reaction given is:
\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl(g) \]

### Step 2: Identify the change in enthalpy (ΔH)
The change in enthalpy for the reaction is given as:
\[ \Delta H^\circ = -185 \text{ kJ} \]

### Step 3: Scale the reaction for 2 moles
Since we are considering the reaction where 2 moles of \( H_2 \) react with 2 moles of \( Cl_2 \) to produce 4 moles of \( HCl \), we need to multiply the entire reaction by 2:
\[ 2 H_2(g) + 2 Cl_2(g) \rightarrow 4 HCl(g) \]

### Step 4: Calculate the new ΔH for the scaled reaction
When we multiply the reaction by 2, we also multiply ΔH by 2:
\[ \Delta H^\circ = 2 \times (-185 \text{ kJ}) = -370 \text{ kJ} \]

### Step 5: Calculate ΔNG (change in the number of moles of gas)
ΔNG is calculated as:
\[ \Delta N_G = \text{moles of products} - \text{moles of reactants} \]
- Moles of products (4 moles of \( HCl \)) = 4
- Moles of reactants (2 moles of \( H_2 \) + 2 moles of \( Cl_2 \)) = 4

Thus,
\[ \Delta N_G = 4 - 4 = 0 \]

### Step 6: Use the relationship between ΔH and ΔU
The relationship between ΔH and ΔU is given by the equation:
\[ \Delta H^\circ = \Delta U^\circ + \Delta N_G RT \]
Where:
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = 300 K

Since ΔNG = 0, the equation simplifies to:
\[ \Delta H^\circ = \Delta U^\circ \]

### Step 7: Conclusion
From the above steps, we have:
\[ \Delta H^\circ = -370 \text{ kJ} \]
Thus,
\[ \Delta U^\circ = -370 \text{ kJ} \]

### Final Answer
The change in internal energy (ΔU) for the reaction is:
\[ \Delta U^\circ = -370 \text{ kJ} \]

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