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. `-185kJ`
C. 370 kj
D. `-370 kJ`

Answer

To find the change in internal energy (ΔU) for the reaction given, we can use the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU):

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

Where:
- ΔH is the change in enthalpy,
- ΔU is the change in internal energy,
- Δn_g is the change in the number of moles of gas,
- R is the universal gas constant (approximately 8.314 J/(mol·K)),
- T is the temperature in Kelvin.

### Step 1: Identify the reaction and given values
The reaction is:
\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \]
Given:
- ΔH = -185 kJ (for the formation of 2 moles of HCl from 1 mole of H2 and 1 mole of Cl2)
- Temperature (T) = 300 K

### Step 2: Calculate Δn_g
Δn_g is calculated as:
\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} \]
In this reaction:
- Moles of products (HCl) = 2
- Moles of reactants (H2 + Cl2) = 1 + 1 = 2

Thus,
\[ \Delta n_g = 2 - 2 = 0 \]

### Step 3: Substitute values into the equation
Using the equation:
\[ \Delta H = \Delta U + \Delta n_g RT \]
Substituting the known values:
\[ -185 \text{ kJ} = \Delta U + 0 \]
Since Δn_g is 0, the equation simplifies to:
\[ \Delta H = \Delta U \]

### Step 4: Solve for ΔU
Thus, we have:
\[ \Delta U = -185 \text{ kJ} \]

### Final Answer
\[ \Delta U = -185 \text{ kJ} \]

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