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

Options

A. (a) `-312.5kJ`
B. (b) `-125.03kJ`
C. (c) `-310 kJ`
D. (d) `-156kJ`

Explanation

To find the standard internal energy change (ΔU) for the given reaction at 298 K, we can use the relationship between enthalpy change (ΔH) and internal energy change (ΔU). The formula we will use is:

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

Where:
- ΔH = change in enthalpy
- ΔU = change in internal energy
- ΔN_g = change in the number of moles of gas (products - reactants)
- R = universal gas constant (8.314 J/(mol·K))
- T = temperature in Kelvin (298 K)

### Step 1: Identify the given values
- The reaction is:
\[ OF_2(g) + H_2O(g) \rightarrow 2HF(g) + O_2(g) \]
- Given ΔH = -310 kJ = -310,000 J (since we need to convert kJ to J for consistency)
- R = 8.314 J/(mol·K)
- T = 298 K

### Step 2: Calculate ΔN_g
ΔN_g is calculated as follows:
\[ \Delta N_g = \text{(moles of products)} - \text{(moles of reactants)} \]
From the reaction:
- Moles of products (2HF + O2) = 2 + 1 = 3
- Moles of reactants (OF2 + H2O) = 1 + 1 = 2

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

### Step 3: Substitute values into the equation
Now we can substitute the known values into the equation:
\[ \Delta H = \Delta U + \Delta N_g RT \]
Substituting the values:
\[ -310,000 J = \Delta U + (1)(8.314 J/(mol·K))(298 K) \]

### Step 4: Calculate the term ΔN_g RT
Calculate:
\[ \Delta N_g RT = 1 \times 8.314 \times 298 = 2477.572 J \]

### Step 5: Rearrange the equation to solve for ΔU
Now we can rearrange the equation to solve for ΔU:
\[ \Delta U = -310,000 J - 2477.572 J \]
\[ \Delta U = -312,477.572 J \]

### Step 6: Convert ΔU back to kJ
Convert ΔU back to kJ:
\[ \Delta U = -312.48 kJ \approx -312.5 kJ \]

### Final Answer
Thus, the standard internal energy change (ΔU) for the reaction at 298 K is approximately:
\[ \Delta U \approx -312.5 \text{ kJ} \]