When `NH_(4)NO_(2)(s)` decomposes at373K , it forms `N_(2)(g)` and `H_(2)O(g)`. The `DeltaH` . For the reaction at one atmospheric pressure and 373 K is -223.6 kJ `mol^(-1)`of `NH_(4)NO_(2)(s)` decomposed. What is the value of `DeltaU` for the reaction under the same conditions ? ( Given `R = 8.31 J K^(-1) mol^(-1))`

Explanation

To find the value of ΔU for the decomposition of NH₄NO₂(s) at 373 K, we can use the relationship between ΔH and ΔU given by the equation:

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

Where:
- ΔH = change in enthalpy
- ΔU = change in internal energy
- Δn_g = change in the number of moles of gas
- R = universal gas constant (8.31 J K⁻¹ mol⁻¹)
- T = temperature in Kelvin

### Step 1: Write the balanced chemical equation
The decomposition of ammonium nitrite can be represented as:

\[ NH_4NO_2(s) \rightarrow N_2(g) + 2H_2O(g) \]

### Step 2: Calculate Δn_g
Δn_g is the change in the number of moles of gas. We can determine this from the balanced equation:

- Reactants: 0 moles of gas (since NH₄NO₂ is a solid)
- Products: 1 mole of N₂ + 2 moles of H₂O = 3 moles of gas

Thus,

\[ \Delta n_g = 3 - 0 = 3 \]

### Step 3: Substitute values into the equation
We know:
- ΔH = -223.6 kJ/mol = -223600 J/mol (since 1 kJ = 1000 J)
- R = 8.31 J K⁻¹ mol⁻¹
- T = 373 K

Now we can substitute these values into the equation:

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

Rearranging gives us:

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

### Step 4: Calculate RT
Now we calculate RT:

\[ RT = 8.31 \, \text{J K}^{-1} \text{mol}^{-1} \times 373 \, \text{K} = 3106.03 \, \text{J/mol} \]

### Step 5: Calculate Δn_g RT
Now we calculate Δn_g RT:

\[ \Delta n_g RT = 3 \times 3106.03 \, \text{J/mol} = 9318.09 \, \text{J/mol} \]

### Step 6: Substitute back to find ΔU
Now substituting back into the equation for ΔU:

\[ \Delta U = -223600 \, \text{J/mol} - 9318.09 \, \text{J/mol} \]

\[ \Delta U = -232918.09 \, \text{J/mol} \]

### Step 7: Convert ΔU to kJ
To convert ΔU back to kJ:

\[ \Delta U = -232.92 \, \text{kJ/mol} \]

### Final Answer
Thus, the value of ΔU for the reaction under the same conditions is approximately:

\[ \Delta U \approx -232.92 \, \text{kJ/mol} \]