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Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

\[ 2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \]

Answer: \( K = 6.9 \)

Given:
- \(\Delta G\) for \(\text{NO}_2 = 51.3 \, \text{kJ/mol}\)
- \(\Delta G\) for \(\text{N}_2\text{O}_4 = 99.8 \, \text{kJ/mol}\)

Answer :

the equilibrium constant (K) for the dissociation of dinitrogen tetroxide at 25°C is approximately 1.001.

To calculate the equilibrium constant (K) for the dissociation of dinitrogen tetroxide (N2O4) at 25°C, we can use the relationship between Gibbs free energy (ΔG°), temperature (T), and equilibrium constant (K), which is given by the equation:

[tex]\[ \Delta G° = -RT \ln(K) \][/tex]

Given that the standard Gibbs free energy change (\( \Delta G° \)) for the reaction \( 2NO_2(g) \rightleftharpoons N_2O_4(g) \) is provided, we can use these values to solve for K.

Given data:

- [tex]\( \Delta G° \) for \( NO_2(g) \): \( 51.3 \, \text{kJ/mol} \)[/tex]

- [tex]\( \Delta G° \) for \( N_2O_4(g) \): \( 99.8 \, \text{kJ/mol} \)[/tex]

- [tex]Temperature (\( T \)): \( 25°C \) (which is \( 298.15 \, \text{K} \))[/tex]

First, let's calculate [tex]\( \Delta G° \)[/tex] for the reaction:

[tex]\[ \Delta G°_{\text{reaction}} = \sum \Delta G°_{\text{products}} - \sum \Delta G°_{\text{reactants}} \][/tex]

[tex]\[ \Delta G°_{\text{reaction}} = 99.8 \, \text{kJ/mol} - (2 \times 51.3 \, \text{kJ/mol}) \][/tex]

[tex]\[ \Delta G°_{\text{reaction}} = 99.8 \, \text{kJ/mol} - 102.6 \, \text{kJ/mol} \][/tex]

[tex]\[ \Delta G°_{\text{reaction}} = -2.8 \, \text{kJ/mol} \][/tex]

Now, we can use the equation [tex]\( \Delta G° = -RT \ln(K) \)[/tex] to solve for K:

[tex]\[ -2.8 \, \text{kJ/mol} = - (8.314 \, \text{J/mol} \cdot \text{K}) \times (298.15 \, \text{K}) \cdot \ln(K) \][/tex]

[tex]\[ \ln(K) = \frac{-2.8 \times 10^3 \, \text{J/mol}}{-(8.314 \, \text{J/mol} \cdot \text{K}) \times 298.15 \, \text{K}} \][/tex]

[tex]\[ \ln(K) \approx \frac{-2.8}{-(8.314 \times 298.15)} \][/tex]

[tex]\[ \ln(K) \approx \frac{-2.8}{-2473.1} \][/tex]

[tex]\[ \ln(K) \approx 0.00113 \][/tex]

Now, take the exponential of both sides to solve for K:

[tex]\[ K = e^{\ln(K)} \][/tex]

[tex]\[ K = e^{0.00113} \][/tex]

[tex]\[ K \approx 1.001 \][/tex]

Therefore, the equilibrium constant (K) for the dissociation of dinitrogen tetroxide at 25°C is approximately 1.001.

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