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

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:

Initial:                0.025        0.025

At eqllm:         0.025-x     0.025+2x

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

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

i think it's c. i'm not sure but i think so.