Consider the following reaction as an example: $$\ce{NaNO3(s) <=> NaNO2(s) + 1/2O2(g)}$$ Once this reaction has achieved equilibrium and then if we added pure $\ce{NaNO3}$ to this system, then is it correct to say that it will favour forward reaction?

My Thoughts

I have gone through a lot of questions related to this topic in Chemistry Stack Exchange and all of them have explained in detail that why it will not affect the value of equilibrium constant (as their "concentration" would not change and I fully agree with that), but no one answers the core question. That if we add or remove a pure solid/liquid then why wouldn't "moles" of other compounds in the reaction change which are included in equilibrium expression?

  • $\begingroup$ I meant that let's say we add some moles of nano3 (s) to the above example then some moles of both nano2 and o2 would form by standard reaction stoichiometry. $\endgroup$ Commented Sep 6, 2020 at 6:48
  • $\begingroup$ Then- 1) Wouldn't change in moles of o2 affect reaction quotient? (Keeping Temperature constant, I know 'Kp' wouldn't change but Reaction Quotient 'Q' should change and hence forward reaction should be more favoured (have higher rate than backward reaction) $\endgroup$ Commented Sep 6, 2020 at 6:56
  • $\begingroup$ 2) If what I said above is true, then wouldn't that mean that changing amount of solid also affects equilibrium but not Kp $\endgroup$ Commented Sep 6, 2020 at 6:59
  • $\begingroup$ It isn't concentration but activity $\endgroup$ Commented Sep 6, 2020 at 7:11
  • $\begingroup$ Where is the carbon on the right side of the chemical equation coming from? $\endgroup$
    – MaxW
    Commented Sep 6, 2020 at 7:22

1 Answer 1


I will revive this topic because it also haunted me. The question is:

[OP] Once this reaction has achieved equilibrium and then if we added pure $\ce{NaNO3}$ to this system, then is it correct to say that it will favour forward reaction?

The answer is no, and nothing will happen. I will justify this from a classical thermodynamic framework.

1. Thermodynamic Equilibrium In heterogeneous reactions, like this one, the number of degrees of freedom is given by Gibbs rule for reacting systems \begin{equation} F = 2 - \text{[Num. of phases]} + \text{[Num. of components]} - \text{[Num. of chemical reactions]} \tag{1} \end{equation} We have:

  • Three components: $\ce{NaNO3}$, $\ce{NaNO2}$, and $\ce{O2}$.
  • Three phases: gas from $\ce{O2}$, one solid phase of $\ce{NaNO3}$, and another solid phase of $\ce{NaNO2}$.
  • One chemical reaction.

Hence, Eq. (1) yields $$ F = 2 - 3 + 3 - 1 \to \boxed{F = 1} \tag{2} $$

Thus, the equilibrium state is fixed with one intensive variable. Say say we are in the lab, and we have a beaker of a volume $V$ at a temperature $T$. We put initially some initial amount $n_\ce{NaNO3}$ of sodium nitrate.

The equilibrium constant at temperature $T$ is denoted by $K$, and its value is \begin{equation} K(T) = \exp\left[-\frac{\Delta_\mathrm{r}G^\circ(T)}{RT}\right] \tag{3} \end{equation}

Now we apply the law of mass action \begin{align} K(T) &= \prod_j a_j^{\nu_j} \\ &= \dfrac{a_\ce{NaNO2} \sqrt{a_\ce{O2}}}{a_\ce{NaNO3}} \\ &= \sqrt{a_\ce{O2}} \\ &= \sqrt{\frac{f_\ce{O2}}{f_\ce{O2}^\circ}} \\ &= \sqrt{\dfrac{\phi_\ce{O2}P_\ce{O2}}{P^\circ}} \\ &= \sqrt{\dfrac{P_\ce{O2}}{P^\circ}} \rightarrow \boxed{P_\ce{O2} = K^2(T)P^\circ} \tag{4} \end{align} since:

  • The activities of solid pure phases are one.
  • The gas phase behaves as an ideal gas, so the fugacity coefficient $\phi_\ce{O2} = 1$.
  • The standard pressure is denoted by $P^\circ$.

The gas phase is composed only of oxygen, thus, the amount of oxygen using Eq. (4) is \begin{equation} n_\ce{O2} = \frac{P_\ce{O2}V}{RT} \rightarrow \boxed{n_\ce{O2}(T) = \left[\frac{K^2(T)V}{RT}\right]P^\circ} \tag{5} \end{equation}

There is a little assumption in the calculation, regarding the volume. In this equilibrium state, the two solid phases will take some space of the beaker, so the volume $V$ in Eq. (5) won't be the volume of the beaker, but a little less.

2. Analysis Eq. (5) affirms that for a fixed temperature $T$ in a volume $V$, we can only have this amount of oxygen in an equilibrium state. Initially we had $n_\ce{NaNO3}$ in the beaker, so given an infinite time when the law of mass action will hold, some $n_\ce{NaNO3}$ will be consumed. How much? By stoichiometry, until we have $n_\ce{O2}$ in the gas phase. Once this value is reached, even though we keep adding $n_\ce{NaNO3}$, no more $\ce{O2}$ can be present in the gas phase.

If we take seriously the fact that adding $n_\ce{NaNO3}$ will take some volume of the beaker, then Eq. (5) will still hold, but $n_\ce{O2}$ will change. If $V$ decreases, then $n_\ce{O2}$ must decrease. We have the funny situation that we add reactant and the reaction will proceed in the reverse direction. For practical purposes, the volume of solids are negligible, and the system will not change its state.

  • $\begingroup$ If 0 deg C ice is added to ice+water in equilibrium, it will not suddenly start freezing nor melting. Adding more of pure solid (isobarically, it is not exactly true for isochoric scenario) does not change its chemical potential nor activity, as these are intensive, not extensive properties. $\endgroup$
    – Poutnik
    Commented Mar 17, 2023 at 15:50
  • $\begingroup$ That nothing will happen is NOT CORRECT. Addition of more of a phase in a form that it can react will increase the rate of the forward reaction and provide more sites for the increased rate of the reverse reaction. A living example: boil a small pot of water until the equilibrium VP is met. Then boil some huge pots of water until drops are condensing and evaporating all over the kitchen. At the same T the VP is the same. In solid state reactions the difference can be control or an explosion. $\endgroup$
    – jimchmst
    Commented Apr 16, 2023 at 5:51
  • $\begingroup$ I could be also said that equilibrium depends on intensive properties, like activity, concentration, partial pressure, fugacity. Not on extensive properties like mass or volume. $\endgroup$
    – Poutnik
    Commented Apr 16, 2023 at 7:00
  • $\begingroup$ @Jimchmst Hello Jimchmst. I am not sure that I have understood you case. Your example of VLE is not a case of chemical reaction, but phase transition, so I do not think it applies. Generally, adding a reactant increases the rate of reaction, which we agree. My conclusion applies to this particular question, where only a unique partial pressure can exist at equilbrium. $\endgroup$ Commented Apr 16, 2023 at 23:21
  • $\begingroup$ True. At equilibrium forward and reverse reactions ae equal that is Thermo. What those rates are is kinetics. Adding nitrate in an appropriate form increases the forward rate increased O2 and Nitrite increases the reverse rate to reestablish equilibrium. All this takes time ie kinetics. Phase changes are as much a chemical reaction as a physical reaction; the same rules apply [the First and Second Laws]. The thermo and Phase Rule approach is exquisite but is equilibrium and needs attention to kinetics. $\endgroup$
    – jimchmst
    Commented Aug 16, 2023 at 19:18

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