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107

One publication for you: “Negative pH Does Exist”, K. F. Lim, J. Chem. Educ. 2006, 83, 1465. Quoting the abstract in full: The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books. The article text provides some counterexamples: For example, commercially available ...

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It's certainly possible theoretically. Solve for $\ce{pH < 0}$: $\ce{-log[H+] < 0\\ log[H+] > 0\\ [H+] > 1}$ So, as you said, a solution in which the hydrogen ion concentration exceeds one should theoretically have a negative $\ce{pH}$. That said, at those extremes of concentration, the utility and accuracy of the $\ce{pH}$ scale breaks down ...

52

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 ...

39

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 $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} ... 38 Dissociation obviously increases the number of moles. The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change. There are two cases on which equilibrium depends. These are: Addition of an inert gas at constant volume: When an inert gas is added to the system in equilibrium at constant volume, the total pressure ... 34 Nick's answer is good. Let's add a little maths. Let's take an example dissociation reaction$\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 Any strong acid solution with concentration more than 1 mol/L has the negative pH. Think about any concentrate commonly used strong acid solution such as 3M$\ce{HCl}$, 6M$\ce{HNO3}$. Negative pH is actually very common. 29 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 ... 27 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{... 26 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 I think your question really arises from some confusion about what \Delta G represents. In general, \Delta X for a thermodynamic quantity X is the change of X along some process. You could make it clear by actually writing \Delta G(\text{A}\rightarrow\text{B}) where A and B are before and after states. (We'll note that, in the general case, \Delta ... 21 It is possible to have \mathrm{pH}<0 and you don't need to create any substance. Take a concentrated solution of one of the strong inorganic acids (i.e. one with dissociation constant above 1000 like sulfuric acid) and here you are. 20 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 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 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-}$$When we talk about equilibrium lying "to the left", it means that the educt/reactant is favored, i.e. more \ce{H2O} than \ce{H3O+} or \ce{OH-}. Conversely, an equilibrium that lies "to ... 19 Yes, equilibrium and steady-state are distinct concepts. A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant. Steady-state implies a system that is not ... 18 Probably you are having problems with Le Chatelier's Principle. Suppose you have an equilibrium established between four substances \ce{A}, \ce{B}, \ce{C} and \ce{D}. What would happen if you changed the conditions by increasing the concentration of \ce{A}? According to Le Chatelier, the position of equilibrium will move in such a way as to ... 18 As noted in this previous question, the correct definition of the equilibrium constant K depends on activities. If you are interested in the derivation of the equation \Delta G^\circ = -RT \ln K (which requires "proper" thermodynamics), read Philipp's answer to that question. For a reaction$$0 \longrightarrow \sum_i \nu_i \ce{J}_i$$(this is a fancy ... 18 Yes, every chemical reaction can theoretically be in equilibrium. Every reaction is reversible. See my answer to chem.SE question 43258 for more details. This includes even precipitation reactions and reactions that release gases. Equilibrium isn't just for liquids! Multiphase equilibria exist. The only thing that stops chemical reactions from being "... 17 Since the stratosphere is nowhere near a closed system, a chlorine atom will eventually leave it. Look at the phrasing again: It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere. It is not stated, that it looses its potential, it just leaves the region, where there is sufficient ozone ... 17 A zeroth order reaction does not achieve equilibrium as a zeroth order reaction. Instead the kinetics will go away from zeroth order when we get close to the equilibrium condition. Zeroth order kinetics is an approximation we get to when some aspect of the reaction approaches a saturation condition, for example the reaction mechanism may involve adsorption ... 15 Let us first define the terms needed here. Degree of dissociation (DOD) Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by \alpha.$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$Number of moles ... 15 This source explains it well. It looks like part of class material, but it clearly explains the dimensionlessness of K_{eq}. The resolution of this apparent paradox is that the above equation, while perfectly satisfactory for everyday use, is not technically correct. A more correct version is:$$K_{eq} = \frac{\frac{\lvert B \rvert_{eq}}{\lvert B \... 15 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, ... 15 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{...

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 ...

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When a particular chemical process is at equilibrium, the respective rates of the forward and backward reactions are equal, meaning that no net change in the concentrations of products and reactants occurs and the composition of the reaction mixture remains stable. If $\ce{OH-}$ is added to a solution already at equilibrium, then there will be an excess of ...

14

You have to analyze the reaction mechanism behind the self-ionization of water to understand better what happen. In fact is not strictly self but a matter of "couple-ionization": you know the water atoms have different electronegativity, the more electronegative oxygen attracts the less electronegative hydrogen of another water molecule forming an hydrogen ...

14

Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient? Yes! The way I handle equilibrium reactions treats them as a "competition" between a particular reaction and its exact reverse: $$\ce{\alpha A + \beta B ->[k_1] \gamma C + \delta D} \tag{1}$$  \ce{\gamma C + \delta D ->[k_{-1}] \alpha A + \...

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