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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.

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

Data:
- ΔG° for NOâ‚‚ = 51.3 kJ/mol
- ΔG° for Nâ‚‚Oâ‚„ = 99.8 kJ/mol

Calculate the equilibrium constant, \( K \).

Note: The answer in the textbook says \( K = 6.9 \), but I keep getting \( K = 3.09 \) as my answer.

Answer :

The value of k is 3.09 to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

To calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25°C, we can use the thermodynamic data provided in Appendix G. According to the problem, we are given the delta G values for [tex]NO_2[/tex] and [tex]N_2O_4[/tex] as 51.3 and 99.8, respectively. We can use the formula Delta G = -RTln(K) to find the equilibrium constant. Plugging in the values, we get:
51.3 + 51.3 = 99.8 - 99.8 - RTln(K)
-48.5 = -RTln(K)
ln(K) = 48.5/RT
K = [tex]e^{(48.5/RT)}[/tex]
At 25°C, R = 8.314 J/(mol K) and T = 298 K, so:
K = [tex]e^{(48.5/(8.314*298))}[/tex] = 3.09

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