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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. ...


4

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 ...


4

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-}]}{[...


4

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. ...


3

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 ...


3

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 ...


2

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 <=> ...


2

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 $X$ ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


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 ...


1

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 ...


1

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 ...


1

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 ...


1

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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