Question

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L reaction vessel, what will the equilibrium concentration of N2 be?

Answer 1

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2$NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor  $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and$O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [$N_{2 (g)}$]=0   ;     [$O_{2 (g)}$ ]= 0    ; [$NO_{(g)}$]=0.60M

C: [$N_{2 (g)}$]=+x   ;     [$O_{2 (g)}$ ]= +x    ; [$NO_{(g)}$]=-2x

E: [$N_{2 (g)}$]=0+x   ;     [$O_{2 (g)}$ ]= 0+x   ; [$NO_{(g)}$]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$= $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$= $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$= $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$$= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M