Given at `25^(@)` C
`4NH_(3)(g)+5O_(2)(g)hArr6H_(2)O(l)+4NO(g)`,
`DeltaH^(@)=-1169 kJ mol^(-1)`
The value of `DeltaU^(@)` for this reaction at `25^(@)` C will be about :

Options

A. `-1181.4 kJ mol`
B. `-1194.8 kJ mol^(-1)`
C. `-1156.6 kJ mol^(-1)`
D. `-1144.2 kJ mol^(-1)`

Answer

To find the value of \( \Delta U^{\circ} \) for the given reaction at \( 25^{\circ} C \), we can use the relationship between enthalpy change (\( \Delta H^{\circ} \)) and internal energy change (\( \Delta U^{\circ} \)) given by the equation:

\[ \Delta H^{\circ} = \Delta U^{\circ} + \Delta N_{G} RT \]

Where:
- \( \Delta N_{G} \) is the change in the number of moles of gaseous substances,
- \( R \) is the universal gas constant (approximately \( 8.314 \, \text{J/mol K} \)),
- \( T \) is the temperature in Kelvin.

### Step 1: Identify the reaction and given values
The reaction is:
\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightleftharpoons 6 \text{H}_2O(l) + 4 \text{NO}(g) \]
Given:
- \( \Delta H^{\circ} = -1169 \, \text{kJ/mol} \)
- Temperature \( T = 25^{\circ} C = 298 \, \text{K} \)

### Step 2: Calculate \( \Delta N_{G} \)
To find \( \Delta N_{G} \), we need to calculate the difference between the moles of gaseous products and the moles of gaseous reactants.

- Moles of gaseous products:
- \( 4 \text{NO}(g) \) contributes 4 moles.
- \( 6 \text{H}_2O(l) \) does not contribute as it is a liquid.

Total moles of gaseous products = \( 4 \)

- Moles of gaseous reactants:
- \( 4 \text{NH}_3(g) \) contributes 4 moles.
- \( 5 \text{O}_2(g) \) contributes 5 moles.

Total moles of gaseous reactants = \( 4 + 5 = 9 \)

Now, we calculate \( \Delta N_{G} \):
\[ \Delta N_{G} = \text{Moles of gaseous products} - \text{Moles of gaseous reactants} = 4 - 9 = -5 \]

### Step 3: Substitute values into the equation
Now we can substitute the values into the equation:
\[ \Delta H^{\circ} = \Delta U^{\circ} + \Delta N_{G} RT \]

Substituting the known values:
\[ -1169 \, \text{kJ} = \Delta U^{\circ} + (-5) \cdot (8.314 \, \text{J/mol K}) \cdot (298 \, \text{K}) \]

### Step 4: Calculate \( \Delta N_{G} RT \)
First, we need to calculate \( \Delta N_{G} RT \):
\[ \Delta N_{G} RT = -5 \cdot 8.314 \cdot 298 = -12499.7 \, \text{J} \approx -12.5 \, \text{kJ} \]

### Step 5: Rearranging the equation
Now, substituting this back into the equation:
\[ -1169 \, \text{kJ} = \Delta U^{\circ} - 12.5 \, \text{kJ} \]

Rearranging gives:
\[ \Delta U^{\circ} = -1169 + 12.5 = -1156.5 \, \text{kJ/mol} \]

### Final Answer
Thus, the value of \( \Delta U^{\circ} \) for the reaction at \( 25^{\circ} C \) is approximately:
\[ \Delta U^{\circ} \approx -1156.5 \, \text{kJ/mol} \]

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