55

Dissociation obviously increases the number of moles. The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change. There are two cases on which equilibrium depends. These are: Addition of an inert gas at constant volume: When an inert gas is added to the system in equilibrium at constant volume, the total pressure ...


54

The equilibrium constant for combustion of organic matter in air with oxygen is not small, but extremely large ($K_\mathrm{eq} \gg 1$), as is expected from a reaction that is simultaneously very exothermic and (usually) increases entropy due to the formation of more gaseous molecules than the input oxygen. The major reason carbon-based life can exist at ...


51

Short answer Does it need to be at $25~^\circ\mathrm{C}$? No. $\Delta_\mathrm{r} G^\circ$ can be defined at any temperature you wish to define it at, since the standard state does not prescribe a particular temperature. If you change the temperature, $\Delta_\mathrm{r} G^\circ$ will change. Does $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} ...


43

Nick's answer is good. Let's add a little maths. Let's take an example dissociation reaction $\ce{A<=>B + C}$ for which $K_p=1$. Since $$K_p = \frac{P_B P_C}{P_A}$$ one equilibrium scenario is $P_A =P_B = P_C = 1 \text{ atm}$. The total pressure $P_T = 3 \text{ atm}$. If we add an inert gas at constant pressure, the total pressure cannot change. ...


30

It is very much possible. Let’s say you put 3 moles of $\ce{HCl}$ into 1 mole of water. $\ce{HCl}$, being a strong acid dissociates completely into $\ce{H+}$ and $\ce{Cl-}$ ions as: $$\ce{HCl -> H+ + Cl-}$$ so after complete dissociation, $[\ce{H+}]=3~\mathrm{mol/L}$ (ignoring the very tiny contribution from water itself) By definition, $$\mathrm{...


30

Your realisation is correct and something chemistry teachers try to hammer into their students’ heads time and time again (and yet, the point is still often lost): Catalysts will never change the thermodynamics of a reaction. They only ease the path of the reaction. Forward and backward reactions will be accelerated equivalently. So what is the benefit of ...


25

What we got here is a dynamic equilibrium, not a static one. For a static one the molecules would need to know when to stop reacting or when to react in a certain way, in a dynamic equilibrium you don't need to know. Here's a ELI5 explanation of what's happening: Let's say you've got a garden with an appletree and hundred of apples are on the ground. ...


24

Yes, equilibrium and steady-state are distinct concepts. A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant. Steady-state implies a system that is not ...


23

The derivation cited can help you understand it mathematically. But the answer also turns out to be satisfactory intuitively. Let me try to explain how. The term $RT\ln Q_c$ is derived form the entropy part or $\Delta G=\Delta H-T\Delta S$, and in a slightly abstract way, represents entropy of mixing of the products and the reactants. This is why ...


20

As noted in this previous question, the correct definition of the equilibrium constant $K$ depends on activities. If you are interested in the derivation of the equation $\Delta G^\circ = -RT \ln K$ (which requires "proper" thermodynamics), read Philipp's answer to that question. For a reaction $$0 \longrightarrow \sum_i \nu_i \ce{J}_i$$ (this is a fancy ...


20

Because equilibrium is dynamic not static There are many things in the world of chemistry which could be simpler but are not. Equilibrium is one of them. There are few chemical processes where equilibrium is equivalent to the situation where everything just stops. Equilibrium normally occurs when the rates of processes going forward are the same as the ...


19

Yes, every chemical reaction can theoretically be in equilibrium. Every reaction is reversible. See my answer to chem.SE question 43258 for more details. This includes even precipitation reactions and reactions that release gases. Equilibrium isn't just for liquids! Multiphase equilibria exist. The only thing that stops chemical reactions from being "...


19

Le Chatelier's Principle does not directly say what happens to concentration ratios. Nor does it directly compare conditions after a disturbance with those before. It says what happens only after a disturbance has been applied. Here, the disturbance is adding one mole of A per liter. When equilibrium is re-established you have only an additional one-half ...


19

Yes, "to the left" refers to the left side of an equilibrium expression. Traditionally, the autoionization of water is written as $$\ce{2H2O <=> H3O+ + OH-}$$ When we talk about equilibrium lying "to the left", it means that the educt/reactant is favored, i.e. more $\ce{H2O}$ than $\ce{H3O+}$ or $\ce{OH-}$. Conversely, an equilibrium that lies "to ...


18

It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see goldbook: Equilibrium Constant Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the ...


18

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements The preface of the book says: The approach: These reactions represent important facets of the elements involved but are typically presented as no more than ...


17

Since the stratosphere is nowhere near a closed system, a chlorine atom will eventually leave it. Look at the phrasing again: It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere. It is not stated, that it looses its potential, it just leaves the region, where there is sufficient ozone ...


17

Consider the choices. Since it's acetic acid, you can rule out any pH values that indicate a basic or neutral solution. Knowing that acetic acid is a weak acid and that vinegar is a fairly dilute solution, you can also rule out any pH values that would indicate either a strong acid or a highly concentrated solution. Eliminating those impossible values leaves ...


17

As others have pointed out, it is purely kinetics, but you may still wonder, why. For a reaction to actually occur (in both directions) and thus for an equilibrium to be reached, you need to overcome the activation energy. In the case of the Haber-Bosch process, this involves breaking the highly stable $\ce{N#N}$ triple bond. Even with the catalysts used, ...


17

A zeroth order reaction does not achieve equilibrium as a zeroth order reaction. Instead the kinetics will go away from zeroth order when we get close to the equilibrium condition. Zeroth order kinetics is an approximation we get to when some aspect of the reaction approaches a saturation condition, for example the reaction mechanism may involve adsorption ...


16

Your reasoning is along the right lines, but it is incomplete. enol of B is more stable because of more substituted alkene. The methyl group in acetone does stabilize the carbon-carbon double bond in the enol form for the reasons you suggest. However, the methyl group also stabilizes the carbonyl double bond in the keto form (see here for example, this is ...


16

What can I say about (A) and (D)? When a carbonyl compound and water are present together, an equilibrium is set up between the carbonyl compound and the corresponding gem-diol. Like any equilibrium, this equilibrium can be shifted to one side or the other by factors that stabilize (or destabilize) one side or the other. If water is present, poly-vicinal ...


16

Andselisk correctly identified the law of dilution and the name Ostwald is often connected with it. $$K_\text{dissociation} = \frac{\alpha^2}{1-\alpha}\cdot c$$ However, the degree of dissociation is $\alpha$ and has "no" unit, i.e. dimensionless quantity. Therefore the unit for the equilibrium constant is that of a concentration, in SI that would be $\pu{...


16

Chemical equilibrium is a type of dynamic equilibrium, but not every dynamic equilibrium is a chemical equilibrium. In a chemical equilibrium there is no change on the macroscopic scale. That means that if you look at the system it seems like nothing is happening, but at molecular scale there are reactions going on and the rate of forward reaction = rate of ...


15

Please have a look at this answer of mine. It contains the derivation of the defining formula of the equilibrium constant. \begin{equation} \log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} \ . \end{equation} This formula shows the temperature dependence explicitely, ...


15

The term "active mass" is a historical term. The concept of an equilibrium constant was developed by Cato Maximilian Guldberg and Peter Waage. The Law of Mass Action has also been referred to as the Law of Guldberg and Waage, historically. Guldberg and Waage defined the term "active mass" in the 1867 Études sur les affinitès chimiques. l'on peut ...


15

The textbook is precisely correct. The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures. In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$. One must be very ...


15

A chemical equilibrium concerns chemical reactions. There should be at least a forward- and backward reaction between two species but more complex systems with multiple individual reactions may occur. The important observation is that there is no macroscopic change to the chemical constituents of the system, i.e. the concentrations of all reaction partners ...


15

Now that you've got $K_{eq}$, you need to take a look at the expression for the equilibrium constant, which is $$K_{eq} = \frac{a_{\text{ice}}}{a_{\text{water}}},$$ where $a$ is the activity of each species. For a pure liquid and a pure solid, the activity is defined as 1, so the expression on the right hand side of that equation (usually referred to as the ...


15

For a chemical reaction, $$\Delta G = -R T \ln K_{eq}$$ $$\implies K_{eq} = e^{\frac{-\Delta G}{RT}} $$ Thus, in order for a chemical reaction to have $K_{eq}=1$*, it is necessary to have $\Delta G = 0$ For some chemical reactions**, we can achieve this by adjusting the temperature and/or pressure. Thus it is possible. However, it's unlikely that the ...


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