At a given temperature,the equilibrium constant (K_p) for the decomposition of dinitrogen tetraoxide to nitrogen dioxide is 0.172.If 0.224 atm N₂O₄ is sealed in a flask,what partial pressure of NO₂ will exist at equilibrium?
N₂O₄(g) ⇌ 2NO₂(g)

Question

Quizplus · Accepted Answer

The equilibrium constant $K_p = 0.172$ for the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$. Let $x$ be the change in concentration of $N_2O_4$ at equilibrium, so the change in $NO_2$ concentration is $2x$. The initial concentration of $N_2O_4$ is 0.224 atm, so at equilibrium, $N_2O_4 = 0.224 - x$ and $NO_2 = 2x$. The equilibrium expression is $K_p = \frac{[NO_2]^2}{[N_2O_4]}$. Substituting the equilibrium concentrations gives $0.172 = \frac{(2x)^2}{0.224 - x}$. Solving this equation for $x$ gives the change needed to reach equilibrium. The partial pressure of $NO_2$ at equilibrium is $2x$, which, when calculated correctly, gives a value close to 0.158 atm.

At a Given Temperature,the Equilibrium Constant (Kp)for the Decomposition of Dinitrogen

At a Given Temperature,the Equilibrium Constant (K_p)for the Decomposition of Dinitrogen