Gaseous `N_(2)O` decomposes at 298 K and forms `N_(2)(g)` and `O_(2)(g)`.The `DeltaH` for the reaction at 1 atm pressure and 298 K is `-163.15 kJ`. What is the value of `DeltaU` for the decomposition of 200 g of `N_(2)O` under the same conditions?

Explanation

To solve the problem, we need to find the value of ΔU (change in internal energy) for the decomposition of 200 g of N₂O under the given conditions. We will use the relationship between ΔH (enthalpy change) and ΔU, as well as the ideal gas law.

### Step-by-Step Solution:

1. **Write the balanced chemical equation** for the decomposition of N₂O:
\[ 2 \text{N}_2\text{O}(g) \rightarrow 2 \text{N}_2(g) + \text{O}_2(g) \]

2. **Identify the change in the number of moles of gas (Δn)**:
- On the reactant side, we have 2 moles of N₂O.
- On the product side, we have 2 moles of N₂ and 1 mole of O₂, totaling 3 moles.
- Therefore, Δn = moles of products - moles of reactants = 3 - 2 = 1.

3. **Use the relationship between ΔH and ΔU**:
\[ \Delta H = \Delta U + \Delta n RT \]
Rearranging gives:
\[ \Delta U = \Delta H - \Delta n RT \]

4. **Substitute the known values into the equation**:
- Given ΔH = -163.15 kJ (which is -163150 J for calculation),
- R (universal gas constant) = 8.314 J/(mol·K),
- T = 298 K,
- Δn = 1.

Now, calculate ΔnRT:
\[ \Delta n RT = 1 \times 8.314 \, \text{J/(mol·K)} \times 298 \, \text{K} = 2477.572 \, \text{J} \approx 2.477572 \, \text{kJ} \]

5. **Calculate ΔU**:
\[ \Delta U = -163.15 \, \text{kJ} - 2.477572 \, \text{kJ} = -165.627572 \, \text{kJ} \approx -165.63 \, \text{kJ} \]

6. **Calculate the number of moles of N₂O in 200 g**:
- Molar mass of N₂O = 28 (N) + 16 (O) = 44 g/mol.
- Moles of N₂O = \(\frac{200 \, \text{g}}{44 \, \text{g/mol}} \approx 4.545 \, \text{mol}\).

7. **Determine ΔU for the decomposition of 4.545 moles of N₂O**:
- The ΔU calculated above is for 2 moles of N₂O.
- Therefore, for 1 mole, ΔU = -165.63 kJ / 2 = -82.815 kJ.
- For 4.545 moles:
\[ \Delta U = -82.815 \, \text{kJ} \times 4.545 \approx -376.39 \, \text{kJ} \]

### Final Answer:
The value of ΔU for the decomposition of 200 g of N₂O is approximately **-376.39 kJ**.