Question

the value of kc for the following reaction is 0.630 at 409 K N2O4(g) --> 2NO2(g) if a reaction vessel at that temperature intitially contains 0.0250 M NO2 and 0.0250 M N2O4, what is the concentration of NO2 at equilibrium

Answer 1

Answer: The concentration of nitrogen dioxide at equilibrium is 0.063 M

Explanation:

We are given:

Initial concentration of nitrogen dioxide = 0.0250 M

Initial concentration of dinitrogen tetraoxide = 0.0250 M

For the given chemical equation:

                        $N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial:                0.025        0.025

At eqllm:         0.025-x     0.025+2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

We are given:

$K_c=0.630$

Putting values in above expression, we get:

$0.630=\frac{(0.025+2x)^2}{(0.025-x)}\\\\x=-0.2013,0.019$

Neglecting the negative value of 'x', because concentration cannot be negative

So, equilibrium concentration of nitrogen dioxide = (0.025 + 2x) = [0.025 + 2(0.019)] = 0.063 M

Hence, the concentration of nitrogen dioxide at equilibrium is 0.063 M