Dinitrogen tetroxide (N₂O₄) decomposes to nitrogen dioxide (NO₂).If $\Delta$H $\degree$ =58.02 kJ/mol and $\Delta$S $\degree$ = 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 the ratio of the forward and reverse rate constants. The forward rate constant is given by the Arrhenius equation:k_f = A * exp(-Ea/RT)where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature. The reverse rate constant is given by:k_r = A * exp(-Ea/RT)where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature.At equilibrium, the forward and reverse rate constants are equal, so we can set the two equations equal to each other and solve for T:k_f = k_rA * exp(-Ea/RT) = A * exp(-Ea/RT)exp(-Ea/RT) = 1-Ea/RT = 0T = Ea/RSubstituting the given values into the equation, we get:T = (58.02 kJ/mol) / (8.314 J/mol K) = 6970 KTherefore, the reactants and products are in their standard states at equilibrium at a temperature of 6970 K.

Dinitrogen Tetroxide (N2O4) Decomposes to Nitrogen Dioxide (NO2) Δ\DeltaΔ H °\degree° =5802 KJ/mol And Decomposes

Dinitrogen Tetroxide (N₂O₄) Decomposes to Nitrogen Dioxide (NO₂) $\Delta$ H $\degree$ =5802 KJ/mol And Decomposes