The brown gas NO₂ and the colorless gas N₂O₄ exist in equilibrium, 2NO₂ $\rightleftharpoons$N₂O₄. In an experiment, 0.625 mole of N₂O₄ was introduced into a 5.00 L vessel and was allowed to decompose until equilibrium was reached. The concentration of N₂O₄ at equilibrium was 0.0750 M. Calculate K_c for the reaction.

Question

Quizplus · Accepted Answer

The equilibrium constant $K_c$ is calculated using the equilibrium concentrations of the products and reactants. Given the initial amount of $N_2O_4$ (0.625 moles in a 5.00 L vessel gives an initial concentration of 0.125 M) and its equilibrium concentration (0.0750 M), we can determine the change in concentration ($\Delta = 0.125 M - 0.0750 M = 0.050 M$). This change represents the concentration of $NO_2$ formed (since 2 moles of $NO_2$ are formed for every 1 mole of $N_2O_4$ that decomposes, the concentration of $NO_2$ is $2 \times 0.050 M = 0.100 M$). The equilibrium constant $K_c$ is calculated as $[NO_2]^2 / [N_2O_4] = (0.100)^2 / 0.0750 = 0.0100 / 0.0750 = 0.133$, which rounds to 0.13 or approximately 0.1 due to significant figures, but given the choices and the calculation method, the closest and correct answer by the provided logic is 7.5, indicating a mistake in the calculation explanation. The correct approach to find $K_c$ involves using the equilibrium concentrations directly in the expression $K_c = [NO_2]^2 / [N_2O_4]$, but the correct calculation should directly consider the changes in concentration due to the reaction's stoichiometry and the given data. The error in explanation points to a misunderstanding or misapplication of the equilibrium constant formula and stoichiometry.

The Brown Gas NO2 and the Colorless Gas N2O4 Exist ⇌\rightleftharpoons⇌

The Brown Gas NO₂ and the Colorless Gas N₂O₄ Exist $\rightleftharpoons$