The reaction of nitrogen with hydrogen to make ammonia has `Delta H= -92 kJ`.
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`
What is the value of `Delta U` ( in kJ) if the reaction of correct out at a constant pressure of 40 bar and the volume changes is `-1.25` litre.

Answer

To find the value of \( \Delta U \) for the reaction of nitrogen with hydrogen to form ammonia, we can use the thermodynamic relationship between enthalpy change (\( \Delta H \)) and internal energy change (\( \Delta U \)):

\[ \Delta H = \Delta U + P \Delta V \]

### Step-by-Step Solution:

1. **Identify Given Values:**
- \( \Delta H = -92 \, \text{kJ} \)
- Pressure \( P = 40 \, \text{bar} \)
- Volume change \( \Delta V = -1.25 \, \text{liters} \)

2. **Convert \( \Delta H \) to Joules:**
\[ \Delta H = -92 \, \text{kJ} = -92 \times 10^3 \, \text{J} = -92000 \, \text{J} \]

3. **Convert Pressure from Bar to Atmosphere:**
\[ 1 \, \text{bar} = 0.9869 \, \text{atm} \quad \text{(approximately)} \]
\[ P = 40 \, \text{bar} \times 0.9869 \, \text{atm/bar} \approx 39.476 \, \text{atm} \]

4. **Convert Volume Change from Liters to Cubic Meters (if necessary):**
- However, since we will use the volume in liters directly with pressure in atm, we can keep it as is.

5. **Calculate \( P \Delta V \):**
\[ P \Delta V = 39.476 \, \text{atm} \times (-1.25 \, \text{L}) = -49.345 \, \text{L atm} \]
- To convert \( L \, atm \) to Joules, use the conversion factor \( 1 \, L \, atm = 101.325 \, J \):
\[ P \Delta V = -49.345 \, \text{L atm} \times 101.325 \, \text{J/L atm} \approx -5000 \, \text{J} \]

6. **Substitute Values into the Equation:**
\[ -92000 \, \text{J} = \Delta U - 5000 \, \text{J} \]

7. **Solve for \( \Delta U \):**
\[ \Delta U = -92000 \, \text{J} + 5000 \, \text{J} = -87000 \, \text{J} \]

8. **Convert \( \Delta U \) back to kJ:**
\[ \Delta U = -87000 \, \text{J} = -87 \, \text{kJ} \]

### Final Answer:
\[ \Delta U = -87 \, \text{kJ} \]