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

Step-by-step 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=([NO_2]^2)/([N_2O_4])`

We are given:

`K_c=0.630`

Putting values in above expression, we get:

`0.630=((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