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The amount of pure solid or liquid does not affect equilibrium, having chemical activity independent on amount, formally assigned to 1. If a solvent is in abundance wrt solutes, it is often considered still as approximately pure solvent, e.g water in context of diluted water solutions. The aqueous (aq) compounds definitely are not pure liquid. They are ...


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I admit Karsten Theis has given en excellent answer for OP's question. However, I'd like to point out that this could be solve without getting confused by $\mathrm{p}K_\mathrm{b}$, which is common with novices when using the Henderson–Hasselbalch equation for buffers. The equation is derived by dissociation of weak acid ($\ce{HA}$): $$\ce{HA + H2O <=> ...


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Buffer equation The Henderson equation for buffers is: $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$ $\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ add up to 14, as do $\mathrm{pH}$ and $\mathrm{pOH}$. So the expression for $\mathrm{pOH}$ is: $$\mathrm{14 - pOH = 14} - \mathrm{p}K_\mathrm{b} + \log{\frac{[\ce{A-}]}{[...


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A is wrong. AgBr is one of the least soluble compounds of Ag. Why don't you have a look in the table of solubility products to be convinced. Further more, AgBr will never react with H2O to produce a strong acid like HBr and a base like AgOH. The reaction goes the other way round. Strong acids react with hydroxydes to produce a salt and water some. C is ...


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The reaction picture as presented has two tick marks (') centered above/below the $\ce{<=>}$ sign. The starting salt is basic formed by the action of a strong base (NaOH) on weak acetic acid, hence the forward reaction has the tick mark moved off to the right. However, upon the addition of the strong acid HCl: $\ce{HCl (aq) <=> H+ (aq) + Cl- (...


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I got the answer, equilibrium favors the right side because (Cd) is a soft metal and (S) is soft too, so it binds better to (Cd) than (O), in accordance with the HSAB theory.


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Yes, you are correct: an unsaturated solution is not in equilibrium. Equilibrium is the state when rate of dissolution is equal to rate of crystallisation. But in an unsaturated solution there is only dissolution of salt in the water and no crystallisation back to salt.


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You might view the relevant principle as follows: components whose activity changes during a reaction tend to put a stop to a reaction before they are completely consumed. Components with constant activity (independent of the amount of other reactants), on the other hand, can be completely consumed (exhausted): they can be limiting reagents, provided they ...


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I agree with Andrew that it depends on whether you define spontaneity based on $\Delta G^\circ <0$ or $\Delta G < 0$. Typically, freshman chemistry books use the former. However, I've never liked equating spontaneity with the sign of $\Delta G^\circ$, prefering to instead use the sign of $\Delta G^\circ$ as an indicator of whether reactants or ...


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I think it is best to start with a real-world example of a spontaneous reaction as cited in electrochemistry, namely, as occurring in an electrochemical (or battery) cell. In particular, comments from Wikipedia: A spontaneous electrochemical reaction (change in Gibbs free energy less than zero) can be used to generate an electric current in ...


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In general it is necessary to consider any entropy changes in determining whether a system is at equilibrium or if a spontaneous change will occur. As there must be an increase in entropy in actual processes then $dS_{system}+dS_{surr}=dS_{irrev} \ge 0$. By using the first law with the last expression and after several steps, we find that in an ...


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It sounds like your confusion arises from not making a distinction between $\Delta G$ and $\Delta G^\circ$ when describing a reaction as spontaneous or not. The $\Delta G^\circ$ is the free energy change for the reaction at the defined "standard" conditions of 1 M solute concentrations and/or 1 bar gas partial pressures of both the reactants and products. ...


1

This question shows that you have probably not really understood what the free enthalpy (or Gibbs energy, or free energy) is. I will try to explain it qualitatively without too much thermodynamics. Let's go ! The origin of the Gibbs energy is coming from Gibbs' reflexions on the spontaneity of chemical reactions. He was trying to find a potential energy ...


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1) Let us calculate first the $\ce{NH_4F}$ problem. The ion $\ce{F^-}$ reacts partially with water according to $$\ce{F^- + H_2O <=> HF + OH^-}$$ The equilibrium constant $K_b$ of this equilibrium is defined by $$K_b = \frac{[HF][OH^-]}{[F^-]}$$ The numerical value of $K_b$ is needed, but it is not tabulated. However it can be obtained from the ...


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WRONG SOLUTION: OP clarified that there are two separate solutions, not a mixture of the salts. Assuming: (1) that concentrations can be used instead of activities (bad assumption...) (2) The concentration of the ions $\ce{H2F-}$ and $\ce{F-}$ is so large that the final concentrations will be the same as the initial concentrations. (3) Since HF is a ...


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From your comments it seems that you are looking for an approximation of a log. I wish you clarified that in the main question without mentioning calculators. It seemed you just wanted to avoid a calculator for some unknown reasons. As Poutnik states, anyone who can post here, will certainly have access to computers and hence the ability to calculate logs. ...


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Padé Approximation for ln(1+x) provides very interesting trade off between simplicity and accuracy ( See also Wikipedia - Padé approximant ): $$P\{ \ln( 1+x ) \} = \frac{x(6+x)}{6+4x}$$ $\ln(1) = 0$, $\ln(2) = 0.7$, $e_\mathrm{max} = 0.00685$, $e_\mathrm{max, rel} \lt 1\% $, $e_\mathrm{RMS} = 0.00258$ This is already a good and fast ...


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Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well? The rate constant would be high, but that does not mean that the net forward rate is high. If you start without product, the initial rate would be high, but would drop to zero very ...


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I think the simplest way to conceptualize this would be to pretend the heat produced by the reaction is a "product" (it's not really correct to say that, but it may help you understand it). Since the reaction is exothermic, you could rewrite the reaction as: N2 + 3H2 ⟶ 2NH3 + Heat If you raised the temperature, you'd add heat, which is a "product". As per ...


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Using concentrations and not activities, user GRSousaJr presents the exact solution to the problem in equations 1-8. However the simplifications don't really seem appropriate. This is a chemistry problem, not a problem in numerical analysis to solve. The gist is that we can make some simplifications based on the number of significant figures and a knowledge ...


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$K_p$ will not change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you have more gas molecules in the products, so an ...


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Actually, this question is not as bad as the comments may make you think. It's actually quite a sensible question, because when we solve the usual, Schrödinger's equation-based, QM problem of energy states of hydrogen atom (or any other atom), we get eigenstates of the Hamiltonian we inserted into the equation. These solutions should thus be stable for ...


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In other words, is there a scenario where I could end up with two possible [solutions]? No, there is always a single solution. The reaction quotient Q assumes values from zero (no products) to infinity (no reactants). It varies monotonically with the extent of reaction (the x in the ICE table). So there is always a solution (because Q covers the full range ...


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An ICE table is moderately complicated, so it ends up seeming a bit mysterious when you get two solutions. But it's really not that mysterious at all. In fact, this sort of thing is pretty common, since it happens anytime you have two variables related to each other through a square. Let's use a very simple example to illustrate this. Suppose you are ...


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How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions? There are three equations: $$\ce{HCOOH <=> H+ + HCOO-}$$ $$\ce{CH3COOH <=> H+ + CH3COO-}$$ $$\ce{H2O <=> H+ + OH-}$$ Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ...


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At equilibrium, the forward and reverse rates have to be equal. For mass action kinetics, this tells you that the reverse rate constant must be equal to the forward rate constant divided by the equilibrium constant.


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In general I would say 'no', thermodynamics only tells us about starting and ending points, thus you know the $\Delta G^\text{o}$ for the reaction but nothing about its actual mechanism, (since time does not come into thermodynamics). As $K_\mathrm{eq}$ is a ratio of rate constants these can take on many different values and still have the same ratio. To ...


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The key insight here is that the amount added, "$X$", is NOT the typical "$x$" in a standard ICE table. The typical $x$ represents the changes in concentration toward equilibrium. The disturbance $X$ is a perturbation away from equilibrium. The initial state of your ICE table should be the state at which you will begin moving to equilibrium, i.e., with $X$ ...


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Calculation of $K_c$ using initial concentrations $K_c = \dfrac{1.2}{0.125 \times 0.4} = 24l mol^{-1}$ +-------------+-------------+--------------+----------+ | Stage | A(aq) | B(aq) | C(aq) | +-------------+-------------+--------------+----------+ | Initial | 0.125M | 0.4M + X | 1.2M | | Change | -y | ...


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