Dinitrogentetraoxide partially decomposes according to the following equilibrium:
N₂O₄(g) 2NO₂(g)
A 1) 00- L flask is charged with 0.400 mol of N₂O₄. At equilibrium at 373 K, 0.0055 mol of N₂O₄remains. K_eqfor this reaction is .

Question

Dinitrogentetraoxide partially decomposes according to the following equilibrium:
N₂O₄(g)

Dinitrogentetraoxide partially decomposes according to the following equilibrium: N<sub>2</sub>O<sub>4</sub><sub> </sub>(g) 2NO<sub>2</sub><sub> </sub>(g) A 1) 00- L flask is charged with 0.400 mol of N<sub>2</sub>O<sub>4</sub>. At equilibrium at 373 K, 0.0055 mol of N<sub>2</sub>O<sub>4 </sub>remains. K<sub>eq </sub>for this reaction is . A) 2.2 × 10<sup>-</sup><sup> </sup><sup>4</sup><sup> </sup> B) 0.22 C) 0.87 D) 13 E) 0.022

2NO₂(g)
A 1) 00- L flask is charged with 0.400 mol of N₂O₄. At equilibrium at 373 K, 0.0055 mol of N₂O₄remains. K_eqfor this reaction is .

Quizplus · Accepted Answer

The equilibrium constant expression is given by:
$$K_{eq}=\frac{[	ext{N}_2	ext{O}_4]}{[	ext{NO}_2]^2}$$

At the start of the reaction, there were 0.400 mol of $	ext{N}_2	ext{O}_4$. At equilibrium, there were 0.0055 mol of $	ext{N}_2	ext{O}_4$, which means that 0.400 - 0.0055 = 0.3945 mol of $	ext{N}_2	ext{O}_4$ has decomposed. Since two moles of $	ext{NO}_2$ are produced for every mole of $	ext{N}_2	ext{O}_4$ that decomposes, the concentration of $	ext{NO}_2$ at equilibrium is 2(0.3945 mol)/0.100 L = 7.89 M.

Substituting the equilibrium concentrations into the equilibrium constant expression gives:
$$K_{eq}=\frac{(0.0055	ext{ mol}/0.100	ext{ L})}{(7.89	ext{ M})^2}=0.87$$

Therefore, the answer is (C).

Dinitrogentetraoxide Partially Decomposes According to the Following Equilibrium