For the dissociation reaction carried out at , the standard Gibbs free energies of formation are giv — Chemical Equilibrium Chemistry Question
Question
For the dissociation reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ carried out at $298\text{ K}$, the standard Gibbs free energies of formation are given as $\Delta G_f^\circ(N_2O_4) = 100\text{ kJ mol}^{-1}$ and $\Delta G_f^\circ(NO_2) = 50\text{ kJ mol}^{-1}$. What is the value of the standard free energy change of the reaction ($\Delta G_{rxn}^\circ$)?
Answer: B
💡 Solution & Explanation
The standard free energy change of the reaction is calculated as $\Delta G_{rxn}^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})$. For this reaction, $\Delta G_{rxn}^\circ = 2 \times \Delta G_f^\circ(NO_2) - 1 \times \Delta G_f^\circ(N_2O_4) = 2(50) - 100 = 100 - 100 = 0\text{ kJ mol}^{-1}$. This implies the equilibrium constant $K=1$.
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