Question

The standard enthalpy and standard entropy of decomposition of \( N_2O_4 \) to \( NO_2 \) are 55.0 kJ mol\(^{-1}\) and 175.0 J/mol respectively. The standard free energy change for this reaction at 25°C in J mol\(^{-1}\) is  (Nearest integer)

Hint:

Remember to always convert units when necessary. In this case, converting the enthalpy from kJ to J helped to maintain consistency in the units for entropy (J/mol).

Answer 1

The decomposition of \(N_2O_4\) to \(NO_2\) is given by: \[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

1. Given Thermodynamic Data:
- Standard enthalpy change: \(\Delta H^\circ = 55.0 \, \text{kJ mol}^{-1} = 55000 \, \text{J mol}^{-1}\)
- Standard entropy change: \(\Delta S^\circ = 175.0 \, \text{J mol}^{-1} \text{K}^{-1}\)
- Temperature: \(T = 25^\circ \text{C} = 298 \, \text{K}\)

2. Gibbs Free Energy Equation:
The standard Gibbs free energy change is calculated using: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

3. Calculation:
Substituting the given values: \[ \Delta G^\circ = 55000 \, \text{J mol}^{-1} - (298 \, \text{K}) (175.0 \, \text{J mol}^{-1} \text{K}^{-1}) \] \[ \Delta G^\circ = 55000 - 52150 = 2850 \, \text{J mol}^{-1} \]

4. Final Result:
The standard free energy change for this reaction at 25°C is \(2850 \, \text{J mol}^{-1}\).

Final Answer:
The final answer is $\boxed{2850}$.

Answer 2

Step 1: Write the given data 
\[ \Delta H^\circ = 55.0 \, \text{kJ mol}^{-1} = 55.0 \times 10^3 \, \text{J mol}^{-1} \] \[ \Delta S^\circ = 175.0 \, \text{J mol}^{-1}\text{K}^{-1} \] \[ T = 25^\circ C = 25 + 273 = 298\, \text{K} \]

Step 2: Apply the Gibbs free energy equation
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] Substitute the given values: \[ \Delta G^\circ = (55.0 \times 10^3) - (298)(175) \]

Step 3: Perform the calculation
\[ \Delta G^\circ = 55,000 - 52,150 = 2,850 \, \text{J mol}^{-1} \]

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Final Answer:

\[ \boxed{\Delta G^\circ = 2850 \, \text{J mol}^{-1}} \]