34

Why does a dissociation reaction shift to the right with the addition of an inert gas?

Todd Minehardt
  • 10,556
  • 25
  • 52
  • 63
pcforgeek
  • 979
  • 5
  • 16
  • 24
  • 1
    http://chemistry.stackexchange.com/questions/15553/applying-the-le-chateliers-principle/15555#15555 – Brinn Belyea Oct 18 '14 at 14:28
  • 1
    Please specify the nature of this dissociation reaction. Does it contain gas phase? – Yomen Atassi Oct 18 '14 at 16:49
  • 1
    @YomenAtassi I am asking in general. I still don't know much about the topic. – pcforgeek Oct 19 '14 at 07:55
  • 8
    Suppose there are two groups of people, your friends and enemies, that are forced to occupy the same room. Most probably there would be a fight which would spread.

    Now, if I add lots of pillows, so that there is no real mixing of your friends and foes, would there be a fight? Yes, but as the only thing to fight is using pillows, there would be no harm done.

    So, basically you see adding something neutral, prevents association, in other words, it increase dissociation. It needs not to be inert gases, any chemical gaseous enough and neutral enough should do the job.

     – Nick Oct 22 '14 at 10:13

2 Answers2

56

Dissociation obviously increases the number of moles.
The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change.

There are two cases on which equilibrium depends. These are:

  • Addition of an inert gas at constant volume:
    When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.

Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.

  • Addition of an inert gas at constant pressure:
    When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Consider the following reaction in equilibrium: $$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$ The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants.

Please read about the Effect of adding an Inert Gas

Harsh jain
  • 1,693
  • 2
  • 11
  • 23
Nick
  • 887
  • 1
  • 10
  • 12
  • 3
    I met a question about adding an inert gas at constant temperature. What will happen in that case? – Aaron John Sabu Mar 01 '17 at 04:33
  • 3
    Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". – Martin - マーチン Sep 10 '18 at 14:47
  • 1
    @AaronJohnSabu Can you tell the complete question?? – Micelle Mar 30 '20 at 14:28
  • 1
    @AnanyaPateriya This was a long long long time back. I'm sorry I can't recall it – Aaron John Sabu Mar 30 '20 at 14:39
  • But will the equilibrium constant change? If not, why? – Aditya Feb 15 '21 at 04:44
  • @Aditya $Q < K$ at constant pressure: https://youtu.be/kBGNRloiZq0?t=886 See Ben's answer below. The $K$ value will change based on temperature and may change based on pressure. So, I assume constant temperature and $K$ does not change. Please read: https://bit.ly/3baepEv

    As I've not visited this topic in quite some time, I have doubts about $K$ in the case of addition at constant volume.

    Also, to clarify other comments, Ideal Gas Law states $PV = nRT$, so note changes from that and "number" of moles is valid vernacular as is number of kilograms albeit they're clunky expression.

     – Nick Feb 15 '21 at 06:26
43

Nick's answer is good. Let's add a little maths.

Let's take an example dissociation reaction $\ce{A<=>B + C}$ for which $K_p=1$. Since

$$K_p = \frac{P_B P_C}{P_A}$$

one equilibrium scenario is $P_A =P_B = P_C = 1 \text{ atm}$. The total pressure $P_T = 3 \text{ atm}$.

If we add an inert gas at constant pressure, the total pressure cannot change. Thus, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ must decrease. If we add some inert gas $\ce{D}$ so that $P_A=P_B=P_C=P_D=\frac{3}{4} \text{ atm}$, the total pressure is still $P_T=3\text{ atm}$. The reaction quotient $Q_p$ is now:

$$Q_p = \frac{P_B P_C}{P_A}=\frac{(\frac{3}{4}\text{ atm})(\frac{3}{4}\text{ atm})}{\frac{3}{4}\text{ atm}}=\frac{3}{4}\text{ atm}<K_p$$

Since $Q_p<K_p$, the equilibrium will shift toward products.

At constant volume, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ can remain constant and equilibrium will be unchanged. Adding inert gas $\ce{D}$ will increase the pressure of the system.

Tyberius
  • 11,686
  • 10
  • 42
  • 86
Ben Norris
  • 42,831
  • 8
  • 123
  • 181