At a given temperature,0.0664 mol N₂O₄(g)is placed in a 1.00 L flask.After reaching equilibrium,the concentration of NO₂(g)is 6.1 × 10^-3 M.What is K_c for the reaction below? N₂O₄(g) 2 NO₂(g)

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:

$K_c = \frac{[NO]^2}{[N_2O_4]}$

Initially, there is only N$_2$O$_4$, so we can write:

$K_c = \frac{(2x)^2}{(0.0664-x)}$

where $x$ is the concentration of N$_2$O$_4$ that reacts to form NO.

At equilibrium, the concentration of N$_2$O$_4$ is:

$[N_2O_4] = 0.0664 - x$

The concentration of NO is given as 6.1 × 10$^{-3}$ M. Since two moles of NO are produced per mole of N$_2$O$_4$ that reacts, we can write:

$[NO] = 2x = 2(6.1	imes10^{-3}) = 1.22	imes10^{-2}$ M

Substituting these values into the equilibrium constant expression:

$K_c = \frac{(1.22	imes10^{-2})^2}{(0.0664-x)}$

We can assume that $x << 0.0664$, so we can approximate:

$K_c \approx \frac{(1.22	imes10^{-2})^2}{0.0664}$

Simplifying and rounding to three significant figures:

$K_c \approx 5.87	imes10^{-4}$

The closest answer choice to this value is C: 5.9 × 10$^{-4}$.

At a Given Temperature,0