What will be the standard internal energy change for the reaction at 298 K ?
`OF_(2(g))+H_(2)O_((g))+2HF_((g)),DeltaH^(@)=-310kJ`

Options

A. `-312.5Kj`
B. `-125.03kJ`
C. `-310 kJ`
D. `-156kJ`

Explanation

To find the standard internal energy change (ΔU°) for the given reaction at 298 K, we can use the relationship between the change in enthalpy (ΔH°) and the change in internal energy (ΔU°) given by the equation:

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

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 (8.314 J/mol·K),
- T is the temperature in Kelvin.

### Step 1: Identify ΔH°
From the question, we have:

\[ \Delta H° = -310 \, \text{kJ} \]

### Step 2: Calculate Δn_g
To find Δn_g, we need to determine the change in the number of moles of gas between the products and reactants.

The reaction is:

\[ OF_2(g) + H_2O(g) \rightarrow 2HF(g) \]

- Moles of gaseous reactants:
- 1 mole of \(OF_2\) + 1 mole of \(H_2O\) = 2 moles of gas

- Moles of gaseous products:
- 2 moles of \(HF\) = 2 moles of gas

Now, we can calculate Δn_g:

\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

### Step 3: Substitute values into the equation
Now we can substitute ΔH°, Δn_g, R, and T into the equation:

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

Since Δn_g = 0:

\[ -310 \, \text{kJ} = \Delta U° + 0 \]

### Step 4: Solve for ΔU°
This simplifies to:

\[ \Delta U° = -310 \, \text{kJ} \]

### Step 5: Convert units if necessary
Since the question is in kilojoules, we can leave the answer as is. However, if we need to express it in joules, we can convert:

\[ \Delta U° = -310 \times 1000 \, \text{J} = -310000 \, \text{J} \]

### Final Answer
Thus, the standard internal energy change for the reaction at 298 K is:

\[ \Delta U° = -310 \, \text{kJ} \]

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