Chemical reactions are invariably associated with the transfer of energy either in the form of hear or light. In the laboratory, heat changes in physical and chemical processes are measured with an instrument called calorimeter. Heat change in the process is calculated as: q= ms `Delta T`, s= Specific heat = `c Delta T`= Heat capacity. Heat of reaction at constant pressure is measured using simple or water calorimeter. `Q_(v)= Delta U`= Internal energy change, `Q_(P) = DeltaH, Q_(P) = Q_(V) + P Delta V and DeltaH = Delta U+ Delta nRT`. The amount of energy released during a chemical change depends on the physical state of reactants and products, the condition of pressure, temperature and volume at which the reaction is carried out. The variation of heat of reaction with temperature and pressure is given by Kirchoff's equation: `(DeltaH_(2) - DeltaH_(1))/(T_(2)-T_(1))= Delta C_(P)` (At constant pressure), `(DeltaU_(2) - DeltaU_(1))/(T_(2)-T_(1)) = DeltaC_(V)` (At constant volume)
The enthalpy change `(DeltaH)` for the reaction `N_(2) (g) + 3H_(2)(g) rarr 2NH_(3)(g)` is `-92.38kJ` at 298 K. The internal energy change `DeltaU` at 298 K is

Options

A. `-92.38 kJ`
B. `-87.42 kJ`
C. 97.34 kJ
D. `-89.9kJ`

Answer

To find the internal energy change (ΔU) for the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \) given that the enthalpy change (ΔH) is -92.38 kJ at 298 K, we can use the relationship between ΔH and ΔU:

\[ \Delta H = \Delta U + \Delta nRT \]

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Enthalpy change, \( \Delta H = -92.38 \, \text{kJ} \)
- Temperature, \( T = 298 \, \text{K} \)
- Universal gas constant, \( R = 8.314 \times 10^{-3} \, \text{kJ/K·mol} \)
- Change in the number of moles of gas, \( \Delta n \)

2. **Calculate Δn**:
- For the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \):
- Moles of gaseous reactants = 1 (from \( N_2 \)) + 3 (from \( H_2 \)) = 4
- Moles of gaseous products = 2 (from \( NH_3 \))
- Therefore, \( \Delta n = \text{Moles of products} - \text{Moles of reactants} = 2 - 4 = -2 \).

3. **Substitute Values into the Equation**:
- Rearranging the equation gives:
\[ \Delta U = \Delta H - \Delta nRT \]
- Substitute the values:
\[ \Delta U = -92.38 \, \text{kJ} - (-2)(8.314 \times 10^{-3} \, \text{kJ/K·mol})(298 \, \text{K}) \]

4. **Calculate the Term \( \Delta nRT \)**:
- Calculate \( \Delta nRT \):
\[ \Delta nRT = -2 \times 8.314 \times 10^{-3} \times 298 \]
\[ = -2 \times 8.314 \times 298 \times 10^{-3} = -4.95 \, \text{kJ} \]

5. **Final Calculation of ΔU**:
- Now substitute back to find ΔU:
\[ \Delta U = -92.38 \, \text{kJ} + 4.95 \, \text{kJ} \]
\[ = -92.38 + 4.95 = -87.43 \, \text{kJ} \]

6. **Conclusion**:
- The internal energy change \( \Delta U \) at 298 K is approximately \( -87.43 \, \text{kJ} \).

### Final Answer:
\[ \Delta U \approx -87.42 \, \text{kJ} \]