# Questions in Equilibrium [closed]

I have two inquiries concerning equilibrium:

1. Does $$K_c = 5.0$$ indicate that the equilibrium mixture contains both products and reactants at an approximately equal amount? I know that $$K_c \approx 1.0$$ is said to be so, however I am unsure of if $$5.0$$ would be considered to be approximately $$1.0$$ in this case.

2. Is a reaction at equilibrium if the forward reaction does not change? I believe it would as an indication of equilibrium is that both the forward and reverse reactions occur at the same rate.

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• I suppose the OP wants confirmation the equilibrium means a dynamic equilibrium, with the equal rate of the opposite reactions. Yes, it is . By "reaction does not change" he probably means the concentrations are constant. – Poutnik Apr 19 at 17:44
• You won't get quicker or better answers if you spam Chemistry.SE with the same questions. This is not a paid service, there is no staff constantly sitting on-line and answering all the questions, and you cannot demand a fast and accurate response (plus you already got one answer in less than an hour, and in addition to that I already voted to reopen your first question). Please visit Help Center to familiarize yourself with the basic principles of this site. – andselisk Apr 20 at 3:11
• If were familiar with the rules, you would've not re-posted the same question just because you were dissatisfied with the first answer given and want a quicker response. I feel pity for you if you see this as aggression or fight, I'm just informing that what you've done wasn't correct; also, as I see it, the answer was as decent as the question itself. – andselisk Apr 20 at 3:27
• @James It is very unusual the rate of reaction being constant. Do you have a particular equilibrium in mind ? – Poutnik Apr 20 at 3:35
• @James correct me if I'm wrong but I choose to disagree. A specific example of an equilibrium refers to an equilibrium example with reactants and products for example the equilibrium formed by iron thiocyanate ions in solution. Without a specific example it is difficult to tell if an equilibrium constant 5.0 corresponds to roughly similar amounts of reactants and products – sab hoque Apr 20 at 4:25

A reaction is in equilibrium,
if the reaction quotient $$Q=K_{\rm c}$$,

For reaction $$\ce{A <=> B}$$ $$K_{\rm c}=5$$ means concentration ratio 1:5. It does not seem to me to be approximately the same concentrations.

The equilibrium constant is the thermodynamical quantity, determining the position of equilibrium. It's value does not say if the reaction is in equilibrium.

About the same concentrations, it means rather the same product of concentrations of reagents and products.

If there is a reaction $$\ce{A + B <=> C + D}$$ with $$K_{\rm c}=1$$ and if concentrations are $$c_{\rm A}=1000c_{\rm B}=100c_{\rm C}=10c_{\rm D}$$ we can hardly speak about equal concentrations at equilibrium. as $$K_c=1=\frac{10\cdot 100}{1000\cdot 1}$$

Additionally, if a reaction is not symmetric in counts of reagents of both sides, equilibrium concentration ratios change with concentrations.

If there is a reaction $$\ce{A <=> 2 B}$$ The same concentrations during equilibrium with $$K_{\rm c}=1$$ occur only for $$c_{\rm A}=1$$. If we consider concentrations of $$\ce{A}$$ and $$\ce{B}$$, the following concentrations are in equilibrium:

• A B
• 0.0001 0.01
• 0.01 0.1
• 1 1
• 100 10
• 10000 100

Yes, you think right, at equilibrium, the rates of the forward and reverse reactions are equal.

If \begin{align} rate_{\rm forward}&=const\\ rate_{\rm backward}&=f([products])\\ \end{align} ("Forward reaction does not change"=has a constant rate, as was explained)

..then as reaction is progressing, the backward reaction rate increases, until it matches the forward rate and reaction is at equilibrium ( regardless of the $$K_{\rm c}$$ value.)

• If you are familiar with the concept of the equilibrium constant, which I expect you are, you would know that if Kc, which is the denotation for the equilibrium constant, is approximately 1.0, the reaction is said to be at equilibrium. The issue arises to what is meant by "approximately". This is in fact what my question is asking. – James Apr 19 at 23:21
• @James The Kc value and reaction being at equilibrium are 2 independent things. Kc value is matter of reaction thermodynamics, describing the position of the equilibrium state with minimal value of Gibbs energy. Being at equilibrium is matter of reaction kinetics. The Kc value does not say if the reaction already reached equilibrium or not. – Poutnik Apr 20 at 3:28
• @James A reaction can be in equilibrium for any value of Kc, e.g $10^{52}$ – Poutnik Apr 20 at 5:12