For the reaction , `FeCO_(3)(s)rarrFeO(s)+CO_(2)(g),DeltaH=82.8kJ` at `25^(@)C`, what is `(DeltaE " or "DeltaU)` at `25^(@)C`?

Options

A. 82.8 kJ
B. 80.32 kJ
C. `-2394.77 kJ`
D. `-3274.75 kJ`

Explanation

To find the change in internal energy (ΔE or ΔU) for the reaction:

\[ \text{FeCO}_3(s) \rightarrow \text{FeO}(s) + \text{CO}_2(g) \]

with the given enthalpy change (ΔH) of 82.8 kJ at 25°C, we can use the following relationship:

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

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

### Step-by-Step Solution:

1. **Identify the gaseous species:**
- In the products, we have 1 mole of CO₂ (g).
- In the reactants, there are no gaseous products (only solid FeCO₃).
- Therefore, ΔN_g = moles of gaseous products - moles of gaseous reactants = 1 - 0 = 1.

2. **Substitute values into the equation:**
- We know ΔH = 82.8 kJ.
- Convert ΔH to Joules: 82.8 kJ = 82,800 J.
- Substitute ΔN_g, R, and T into the equation:
\[ \Delta H = \Delta U + \Delta N_g RT \]
\[ 82,800 J = \Delta U + (1)(8.314 \, \text{J/(mol·K)})(298 \, \text{K}) \]

3. **Calculate RT:**
- Calculate RT:
\[ RT = 8.314 \, \text{J/(mol·K)} \times 298 \, \text{K} = 2477.572 \, \text{J} \]

4. **Substitute RT back into the equation:**
\[ 82,800 J = \Delta U + 2477.572 J \]

5. **Solve for ΔU:**
\[ \Delta U = 82,800 J - 2477.572 J \]
\[ \Delta U = 80,322.428 J \]

6. **Convert ΔU back to kilojoules:**
\[ \Delta U = \frac{80,322.428 \, \text{J}}{1000} = 80.322 \, \text{kJ} \]

### Final Answer:
\[ \Delta U \approx 80.32 \, \text{kJ} \]