Can someone help me with this question?

Mar 4, 2018

The equilibrium will shift to the left.

Explanation:

The idea here is that the value of the equilibrium constant tells you where the position of the equilibrium lies. In other words, the value of the equilibrium constant tells you which reaction, i.e. the forward reaction or the reverse reaction, is favored at the given temperature.

However, the direction in which the equilibrium shifts is given by the result of the comparison between the reaction quotient, ${Q}_{c}$, and the equilibrium constant, ${K}_{c}$.

More specifically, you have

• ${Q}_{c} < {K}_{c} \implies$ the equilibrium will shift to the right, i.e. the forward reaction will be favored
• ${Q}_{c} = {K}_{c} \implies$ the reaction is at equilibrium, i.e. the forward reaction and the reverse reaction take place at the same rate, so the equilibrium will not shift
• ${Q}_{c} > {K}_{c} \implies$ the equilibrium will shift to the left, i.e. the reverse reaction will be favored

In your case, you know that at ${500}^{\circ} \text{C}$, this equilibrium reaction

${\text{N"_ 2"O"_ (4(g)) rightleftharpoons color(red)(2)"NO}}_{2 \left(g\right)}$

has

K_c = (["NO"_ 2]_ "equilibrium"^color(red)(2))/(["N"_ 2"O"_ 4]_ "equilibrium")

and

Q_c = (["NO"_ 2]_ "initial"^color(red)(2))/(["N"_ 2"O"_ 4]_ "initial")

So all you have to do here is to calculate the value of the reaction quotient and compare it to the value of the equilibrium constant.

${Q}_{c} = {\left(2.51\right)}^{\textcolor{red}{2}} / 6.13 = 1.03$

Since

$1.03 > 0.674$

you can say that the equilibrium will shift to the left, meaning that the reaction will consume nitrogen dioxide and produce dinitrogen tetroxide.

So when the reaction reaches equilibrium, you should expect the reaction vessel to contain

["NO"_2] < "2.51 M"

["N"_ 2"O"_ 4] > "6.13 M"