Dinitrogen tetroxide (N₂O₄)decomposes to nitrogen dioxide (NO₂).If $\Delta$H $\circ$ = 58.02 kJ/mol and $\Delta$S $\circ$ = 176.1 J/mol . K,at what temperature are reactants and products in their standard states at equilibrium?

Question

Quizplus · Accepted Answer

The equilibrium constant is given by:K = exp(-ΔG°/RT)where ΔG° is the standard free energy change, R is the gas constant, and T is the temperature.We are given that ΔG° = 58.02 kJ/mol and R = 8.314 J/mol K. Substituting these values into the equation, we get:K = exp(-58020 J/mol / (8.314 J/mol K * T))We are also given that the equilibrium constant is 176.1 J/mol K. Substituting this value into the equation, we get:176.1 J/mol K = exp(-58020 J/mol / (8.314 J/mol K * T))Solving for T, we get:T = 56.3 KTherefore, the correct answer is A.

Dinitrogen Tetroxide (N2O4)decomposes to Nitrogen Dioxide (NO2) Δ\DeltaΔ H ∘\circ∘ = 58

Dinitrogen Tetroxide (N₂O₄)decomposes to Nitrogen Dioxide (NO₂) $\Delta$ H^$\circ$ = 58