Thermal Physics: Interpretation of equilibrium constant

[WWCY]

For a reaction defined as such,
$$A\rightleftharpoons B$$
the equilibrium constant $K$ is defined by $K = p_B / p_A$, with $p$ denoting the partial pressure (edit: at equilibrium). However, if $K<<1$, which implies $p_A >> p_B$, it is said that the backwards reaction dominates and that the container will be filled mainly with $A$. $K>>1$ then implies the opposite, and the container will be filled mainly with $B$.

Why is this the case? For example, if I had a container filled only with $A$, which means $K = 0 < 1$, how does any backward reaction dominate? Shouldn't the "A converts to B" reaction dominate?

Thanks in advance!

 

[DrClaude]

$K$ is a constant that is set independently from the partial pressures. It is calculated by considering the change in Gibbs free energy for the reaction.

If $p_B = 0$, then the system is necessarily out of equilibrium and the reaction will proceed from left to right until $p_B / p_A = K$.

 

[WWCY]

$K$ is a constant that is set independently from the partial pressures. It is calculated by considering the change in Gibbs free energy for the reaction.

If $p_B = 0$, then the system is necessarily out of equilibrium and the reaction will proceed from left to right until $p_B / p_A = K$.

Does this also mean that if K >> 1, the reaction will proceed such that the container at equilibrium will have $p_B >> p_A$?

Thanks!

 

[DrClaude]

Does this also mean that if K >> 1, the reaction will proceed such that the container at equilibrium will have $p_B >> p_A$?

Thanks!

Yes. $K>1$ will lead to an excess of B over A, while $K<1$ will lead to an excess of A over B.

 

[DrClaude]

I should add that the values of $K$ (or $\Delta G$) apply to the reaction as written, proceeding from left to write. If you write the equation the other way around, $B \rightleftharpoons A$, then $K \rightarrow 1/K$ and $\Delta G \rightarrow -\Delta G$.

 

[Chestermiller]

If P is the total pressure, and you start out with all A, then at equilibrium, the partial pressure of A will drop to $p_A$ and the partial pressure of B will rise to $p_B=(P-p_A)$. So you would have: $$\frac{p_B}{p_A}=\frac{(P-p_A)}{p_A}=K$$This means that, at equilibrium, $$p_A=\frac{P}{(1+K)}$$and $$p_B=\frac{PK}{(1+K)}$$

 

[WWCY]

Cheers, that cleared things up a lot!