# Doubt in heterogenous equilibria [closed]

Let us take the following equilibrium: $$\ce{NH4HS(s) <=> NH3(g) + H2S(g)}$$

I have been taught that since $$\ce{NH4HS}$$ is in the solid state, its concentration cant be taken and hence we write $$K_{\mathrm{eq}}=[\ce{NH3}][\ce{H2S}]$$

This means that according to Le-Chatelier's Principle, alteration in $$[\ce{NH3}],[\ce{H2S}]$$ results in equilibrium is disturbed. However, alteration in $$[\ce{NH4HS}]$$ won't disturb the equilibrium. This seems a bit confusing to me.

Logically, if we had large quantity of $$\ce{NH4HS}$$, large quantities of $$\ce{NH3}$$,$$\ce{H2S}$$ would be produced. However, the equation tells us that $$[\ce{NH4HS}]$$ won't matter at all. Please help resolve my confusion.

• Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. Commented Nov 17, 2022 at 6:12
• The change in $\ce{[NH3]}$ and $\ce{[H2S]}$ when $[\ce{NH4HS}]$ is added is negligible compared to the change in $[\ce{NH4HS}]$ when $\ce{[NH3]}$ and $\ce{[H2S]}$ are added. Commented Nov 17, 2022 at 6:34
• Large quantity of [X] is nonsense. Either large quantity of X, either high [X], either high concentration of X. Commented Nov 17, 2022 at 7:45
• For WHat reason my question is being downvoted? I have asked for a simple physical chemistry concept along with my undertsanding. Commented Nov 17, 2022 at 18:22
• If no downvote reason is provided, there can be often applied the default one, revealed at the downvote bottom ( The question does not show any research effort; it is unclear or not useful. ). // Always try to find answers on your own in offline and online resources, before asking. It will give the user more than if one just waits to receive the answer. Questions that look like it has not been done are frown upon on StackExchange site network. Commented Nov 18, 2022 at 8:07

• A different amount of solid substance does not shift the equilibrium.
• A different volume ( but the same partial pressures) of gaseous phase does not shift the equilibrium.

Because these are extensive properties of systems.
An equilibrium is based on intensive properties, like concentration, activity, partial pressure, molar fraction or chemical potential.

However, alteration in [$$\ce{NH4HS}$$] won't disturb the equilibrium.

The point is, adding $$\ce{NH4HS(s)}$$ does not mean alteration in [$$\ce{NH4HS(s)}$$], similarly as more of the same gaseous phase does not mean alteration in [$$\ce{NH3(g)}$$] nor [$$\ce{H2S(g)}$$].

At the reaction equilibrium, there is minimum of the Gibbs energy

$$G=H-TS=U + pV - TS,\tag{1}$$

therefore $$\mathrm{d}G = 0 \tag{2}$$.

There is defined the quantity chemical potential $$\mu$$ of a substance as the partial derivative of Gibbs energy of the system per molar amount of such a substance.

$$\mu_i = \left(\frac{\partial G}{\partial n_i} \right)_{\text{const }T,p,n_j; j \ne i}\tag{3}$$

At constant temperature and pressure

$$\mathrm{d}G = \sum_i{\mu_i \cdot \mathrm{d}n_i } \tag{4}$$

$$\mathrm{d}G = \mu_{\ce{NH4HS(s)}}\cdot \mathrm{d}n_{\ce{NH4HS(s)}} + \mu_{\ce{NH3(g)}}\cdot \mathrm{d}n_{\ce{NH3(g)}} + \mu_{\ce{H2S(g)}}\cdot \mathrm{d}n_{\ce{H2S(g)}}\\ =\left(\mu_{\ce{NH4HS(s)}} - \mu_{\ce{NH3(g)}} - \mu_{\ce{H2S(g)}}\right)\mathrm{d}n_{\ce{NH4HS(s)}}\tag{5}$$

and due (2):

$$\mu_{\ce{NH4HS(s)}}= \mu_{\ce{NH3(g)}} + \mu_{\ce{H2S(g)}} \tag{6}$$

Now, $$\mu$$ of solid substances is practically independent on their amounts, if we neglect changes of pressure of otherwise gaseous systems.

$$\frac{{\mathrm{d}\mu_{\ce{NH4HS(s)}}} }{ {\mathrm{d}n_{\ce{NH4HS(s)}}}} \approx 0 \tag{7}$$

Therefore, $$\mu$$ of $$\ce{NH3(g)}$$ and $$\ce{H2S(g)}$$ and therefore their partial pressures or concentrations are (practically) independent on the amount of $$\ce{NH4HS(s)}$$, as for ideal gases,

$$\mu = \mu^{\circ} + RT\ln{\frac{p}{p_0}}\tag{8}$$

Therefore, the amount of the solid component does not have (in the first approximation) impact on the equilibrium.

Note that (molar) concentrations are not much used for gases. The formal relation of concentration and partial pressure is:

$$n_i=\frac{p_iV}{RT} \implies \frac{n_i}{V}=\frac{p_i}{RT} \implies c_i = \frac{p_i}{RT} \tag{9}$$

Feedback response:

For a pure solid substance, its $$\mu$$ is a parameter characteristic for the substance itself, not for its amount. By other words, $$\ce{NH4HS(s)}$$ does not have bigger thermodynamic tendency to decompose if there is a bigger amount of it.

It is analogical as having bigger volume of the gaseous phase, but with the same partial pressures of $$\ce{H2S(g)}$$ and $$\ce{NH3(g)}$$, therefore having their respective $$\mu$$ independent on the gas phase volume.

There would not be created more of $$\ce{NH4HS(s)}$$ if there is doubled the gas phase volume, keeping partial and total pressures. OTOH, if there is increased their partial pressure, their $$\mu$$ would increase and the there would be formed more $$\ce{NH4HS(s)}$$, until $$\mu$$ of substances get balanced again.

• So our chemical potential of a compound is analogous to something like density which is also independant of initial mass/concentration. Am I right? Commented Nov 17, 2022 at 11:22
• See the answer update. Commented Nov 17, 2022 at 11:55
• So, adding more $\ce{NH4HS}$(s) will not produce more $\ce{NH3}$(g) and more $\ce{H2S}$(g) Commented Nov 17, 2022 at 15:37
• @Maurice Assuming the system is already in equilibrium and addition of solid causes zero (p=const) or negligible shift of partial pressures, it will not. Different initial amounts will lead to the same equilibrium partial pressures. The key is Eq. no (6). Commented Nov 17, 2022 at 15:41
• @Poutnik. So you agree with my interpretation. Commented Nov 17, 2022 at 17:21

I have been taught that since $$\ce{NH4HS}$$ is in the solid state, its concentration cant be taken and hence we write $$K_{\mathrm{eq}}=[\ce{NH3}][\ce{H2S}]$$

The other way to think about it is that the concentrations in the equilibrium constant expressions are relative to a standard state (so they are dimensionless, without units). The standard state for a solid sample is the pure solid sample, so it is always 1.

This means that according to Le-Chatelier's Principle, alteration in $$[\ce{NH3}],[\ce{H2S}]$$ results in equilibrium is disturbed. However, alteration in $$[\ce{NH4HS}]$$ won't disturb the equilibrium. This seems a bit confusing to me.

Here is a home chemistry example. If you have a saturated salt solution, will adding more solid pure salt change the concentration in solution? No, it will not. However, if you lower the concentration of salt in the solution (by adding more water), the equilibrium will be disturbed, and more solid salt will dissolve.

Logically, if we had large quantity of $$\ce{NH4HS}$$, large quantities of $$\ce{NH3}$$,$$\ce{H2S}$$ would be produced. However, the equation tells us that $$[\ce{NH4HS}]$$ won't matter at all. Please help resolve my confusion.

Quantity and concentration are two different things (the first is extensive, and the second is not). Let's say you have some $$\ce{NH4HS}$$ in a closed bottle. It makes a certain amount of ammonia and dihydrogen sulfide, according to the equilibrium expression. If you want to make more of these gases, you could set up more bottles. This shows you that you can scale this up. The concentration in each bottle would still be the same, but you would have a larger amount.

Alternatively and less complicated, you could switch to a larger bottle, or you could continously pump out the gases from your bottle to make more (higher amount of substance) of the gas species. However, the concentration of the gases would still be the same (if it were higher, you would produce some $$\ce{NH4HS}$$ again.

A PURE solid or a PURE liquid has a constant chemical composition and constant chemical activity. There is a dependence on crystal or droplet size when either become very small and most of the molecules are exposed and not in contact with others hence the interest in nanochemistry. The amount does have an effect on kinetics, the time it takes to reach equilibrium, and there must be enough to establish equilibrium. You seem to think that its concentration, activity, is not considered. It is definitely considered! It is a constant and therefore can be defined; it is convenient to define it as 1.. If the solid or liquid is impure its activity must be redefined, usually as the mole fraction or as molality [moles per constant mass of solvent].

Instead of worrying about solid state chemical reactions take a step back and investigate simple phase changes such as water-ice where both activities are 1. or water-water vapor where the activity of the water can be considered 1. and the activity of the vapor is its vapor pressure. [The second case is more complicated since the activity of the water is a function of temperature because of the higher vapor pressure with higher T but you can redefine it as 1. at each temperature.] Once understood then try some simple reactions such as NH4HS decomposition