# Question #145f0

Feb 11, 2016

Will shift to the left side.

#### Explanation:

The equilibrium in question is:

$N {H}_{4} C l \left(s\right) r i g h t \le f t h a r p \infty n s N {H}_{3} \left(g\right) + H C l \left(g\right)$

According to Le Châtelier's principle, when a gaseous or aqueous component of a system at equilibrium is added to the system, the equilibrium will shift to the side that will minimize the amount of the component added.

Since $N {H}_{3}$ is a gas and it is a product, the equilibrium will shift to the reactant side (left side).

From the pressure point of view, Le Châtelier's principle also states that if the pressure is increased by adding a gaseous component of the system, the equilibrium will readjust by decreasing the pressure and the result is as discussed above.

However, adding $N {H}_{4} C l$, and since it is at the solid state, the equilibrium will not be affected.

Feb 12, 2016

Yes, the equilibrium will shift left.

#### Explanation:

As you know, chemical equilibria are governed by Le Chaterlier's Principle, which states that a system at equilibrium will react to any stress placed on that equilibrium in such a way a to reduce that stress.

No,w you're dealing with a heterogeneous equilibrium, which involves substances in more than one phase. The most important thing to keep in mind about this type of equilibrium is that the concentrations of solids and pure liquids are presumed constant.

In other words, changing the concentration of a solid will Not affect the position of the equilibrium whatsoever.

Now, your equilibrium looks like this

${\text{NH"_4"Cl"_text((s]) rightleftharpoons "NH"_text(3(g]) + "HCl}}_{\textrm{\left(g\right]}}$

The equilibrium constant expressed in terms of partial pressures for this reaction takes the form

${K}_{p} = \left(\text{NH"_3) * ("HCl}\right)$

Similarly, you have

${K}_{c} = \left[\text{NH"_3] * ["HCl}\right]$

Remember, the solid is not included in the expression for the equilibrium constant.

So, this tells you that a given temperature, increasing the partial pressure of ammonia will cause the equilibrium to shift to the left, since the reverse reaction is the one that consumes ammonia, and thus reduces its partial pressure.

${\text{NH"_4"Cl"_text((s]) rightleftharpoons "NH"_text(3(g]) + "HCl}}_{\textrm{\left(g\right]}}$

$\textcolor{w h i t e}{a a a a} \stackrel{\textcolor{red}{\leftarrow}}{\textcolor{w h i t e}{a a a} \textcolor{g r e e n}{\text{shift to the left}} \textcolor{w h i t e}{a a a a}}$

Therefore, you can expect the reaction to consume gaseous ammonia and hydrogen chloride and produce solid ammonium chloride.

It is very important to realize that the opposite does not take place! Increasing or decreasing the amount of ammonium chloride will not change the position of the equilibrium!

The position of the equilibrium will only change when a concentration or a partial pressure that appears in the expression of the equilibrium constant changes.

You can read more on that here and here.