It is my understanding that when ∆Stot / ∆Suniv = 0 then the reaction does not occur, the position of equilibrium is not shifted in a specific direction. However, does this imply that [reactants] = [products]?

The question 1e part iii asks "What can you understand about the the equilibrium when ∆Stot = 0". The answer to the question is "Kc = 1, or reaction balanced in middle, or no tendency to go in either direction"

To me, this only makes sense if the starting concentrations of the reactants and products are the same. However, at no point is it mentioned that the starting concentrations of reactants/products are equal.


1 Answer 1


All thermodynamics boils down to looking at $\Delta S_\mathrm{univ}$. $\Delta S_\mathrm{univ}$ is the sum of $\Delta S$ for your system and $\Delta S$ for everything else. We measure $\Delta S$ for the system directly and we measure $\Delta S$ for everything else via $-\frac{\Delta H}{T}$. When these two are equal, notice that $\Delta G = \Delta H - T\Delta S = 0$. Then given $\Delta G = -RT \ln K$, we conclude that $K = 1$.

The reaction favors reactants as much as products, but that specific balance depends on the reaction and its coefficients, i.e., the equilibrium expression could have powers. Consider: $$\ce{A2 -> 2A}$$

The concentrations of product and reactant are not equal at equilibrium for $K = 1$.

Also, it does not mean that a reaction does not occur. You need to compare the reaction quotient $Q$ against $K$ to determine if there is a reaction or if we are actually at equilibrium.


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