How do the coefficients in an equilibrium question affect the shifts in position of equilibrium?

Let's say I have this equation: $$A_{(g)}\leftrightharpoons 2B_{(g)}$$

According to Le Chatelier's principle, if I decrease the pressure, then more of B will be made in order to increase the pressure again, because the forward reaction produces more moles of gas.

However, what if my $$K_C$$ was very small, so that the concentration of A was much greater than B. A would have way more moles of gas than B, so how can you say that the forward reaction produces more moles of gas? I don't understand what the coefficients of the equation actually represent since you can't say that the ratio of concentration or partial pressure, A:B, is always 1:2.

Could you say that the coefficients just tell you which species is most affected by a change?

• @Poutnik Can I say, in general, that the larger the coefficient in front of a species is, the more it will be affected by an imposed change (e.g. change in pressure or dilution), due to the coefficient becoming a power in Kc calculations? Commented Mar 12, 2023 at 5:33
• Yes, I only really understood it then. I wanted to check my understanding. Sorry if this is not how Stack Exchange works, I'm not familiar with the community yet. Commented Mar 12, 2023 at 6:55
• I don't have enough reputation to comment on the other post you wrote, so I will ask here, if that's okay. So, if I have $3A \leftrightharpoons 2B$ then would the frequency be expressed as $k[A][A][A]=k[A]^3$? I get that k represents that rate is proportional to [A], and the second [A] represents how the number of other 'skaters' also affects rate. So what does the third [A] represent? Commented Mar 12, 2023 at 9:14
• some details can be found here that shows what happens when the pressure changes in terms of the amount of A and B chemistry.stackexchange.com/questions/98770/… Commented Mar 12, 2023 at 9:53

The reaction stoichiometric coefficients do not say which component is most affected by the change, but rather which component pressure/concentration has the biggest impact on the reaction quotient(*)

With temperature change of the equilibrium constant, the component with higher order of power have equlibrium presure/concentration less affected, as it is boosted by the exponent.

The reaction stoichiometric coefficients say nothing about the ratio of pressures nor concentrations of $$\ce{A}$$ and $$\ce{B}$$. They just say 1 molecule of $$\ce{A}$$ forms 2 molecules of $$\ce{B}$$ and vice versa.

But, they determine the exponents in the expression for $$K_c$$ or $$K_p$$.

$$K_c = \frac{{[\ce{B}]}^2}{[\ce{A}]} \tag{1}$$

$$K_p = \frac{({p_{\ce{B}})}^2}{p_{\ce{A}}} \tag{2}$$

Increasing of pressure causes the right side becoming bigger than $$K_p$$ (and $$K_c$$ for gaseous phase) and some $$\ce{B}$$ reacts to form $$\ce{A}$$ to make them equal again. And vice versa if the pressure decreases.

Imagine two times bigger pressure concentration means four times more frequent $$\ce{B-B}$$ collisions and four times faster $$\ce{A}$$ production While $$\ce{B}$$ is produced just two times faster.

Note that the value of $$K_c$$ or $$K_p$$ is irrelevant here, it is covered by the proportionality constants for reaction rates. As the fraction of collisions leading to reaction varies greatly.

See the ice-skating analogy and another question Why are the stoichiometric coefficients the powers in the rate law?

REferring to the linked ice-skating analogy and further elaborating it:

The 3rd $$[\ce{A}]$$ represents another skater, colliding with a heap.

The frequency of the collisions of a 2-skater heap and another skater is $$k_3[\ce{H}][\ce{S}] \tag{3}$$

The "concentration" of heaps is $$[\ce{H}]=k_1[\ce{S}][\ce{S}], \tag{4}$$

It gives together:

$$k_3k_1[\ce{S}][\ce{S}][\ce{S}] = k_4[\ce{S}]^3 \tag{5}$$

So, the frequency of collisions of skaters with a 2-skater heap is proportional to the third power of skater "concentration".

But better not to interpret how to divide 3 skaters between the same 2 parts.

Be aware of the note in the other post that it is valid for elementary reactions. Complex reactions with various intermediates have more complex relations in reaction kinetics. Bringing the terms from the equilibrium constants to rate equations is not generally valid, as they do not generally describe the opposite directions of the same elementary reaction.

(*) - the expression for reaction equilibrium constant, but with current values for the reaction, being generally not in equilibrium