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24 votes

Can single molecules of C and O2 react in isolation, and if so how will momentum be conserved?

$\ce{C + O2}$ is awfully complicated, so let's just pretend you've asked this: In a single act of the reaction $\ce{H. + H .-> H2}$, how is momentum conserved? That's a legitimate concern all right....
Ivan Neretin's user avatar
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19 votes

Why is entropy favorable?

Thermodynamics. The second law of thermodynamics states that entropy always increases in an isolated system. This is taken as a fundamental postulate---we simply accept this statement as a fact ...
a-cyclohexane-molecule's user avatar
16 votes
Accepted

Why would there be a non-zero Gibbs energy of mixing for ideal gases?

When we talk about mixing, we usually mean that two components are in different parts of a container before mixing and then they share the entire container after mixing. In other words, the total ...
Karsten's user avatar
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15 votes
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What is the difference between ΔG and ΔrG?

Not for the faint-hearted: There is an excellent, but very mathsy, article here: J. Chem. Educ. 2014, 91, 386 describing the difference. The Gibbs free energy change, $\Delta G$ You are quite right ...
orthocresol's user avatar
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15 votes
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Why proton concentration is divided by 10⁻⁷?

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 ...
andselisk's user avatar
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13 votes
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Why is entropy favorable?

It appears you're looking for an ELI5-style answer, not an elaborate definition. Entropy just happens – as long as the universe isn't frozen solid, things will always be moving around, and that ...
sk29910's user avatar
  • 248
13 votes
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If change in free energy (G) is positive, how do those reactions still occur?

$\Delta G^\circ_m$ is the difference in molar Gibbs free energy between the reagents and products in their standard states (in the case of $\ce{AgI(s)}$, the standard state for the reagent is the pure ...
Buck Thorn's user avatar
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12 votes

Why exactly are standard potentials additive?

Following on Derek's great answer, it is very important to remind that the conventional way we use to add half-cell potentials is a consequence of the conservation of energy. Therefore, we should look ...
truffaut's user avatar
  • 311
11 votes

True or false: "If a reaction has a large negative value of ∆G, then it will be a fast reaction."

There is the Bell–Evans–Polanyi principle stating that "the difference in activation energy between two reactions of the same family is proportional to the difference of their enthalpy of ...
Snijderfrey's user avatar
10 votes

Why is entropy favorable?

Do not think of entropy as 'disorder' as this is misleading, better is that it is a 'measure of disorder' but this is equally vague. It is better to think of entropy as the number of ways that '...
porphyrin's user avatar
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10 votes

Why would there be a non-zero Gibbs energy of mixing for ideal gases?

With your reply you seem to have answered part of your question. I would like to add a bit about the other part. I will have to do it mathematically first, there is no other shortcut, but then we will ...
Metal Storm's user avatar
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8 votes
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Why can't a reaction go to completion?

You alluded to the answer when you mention activation energy. Kinetically the equilibrium constant is $K_e = k_f/k_b$ where $k_f$ and $k_b$ are the forward are reverse reaction rate constants in the ...
porphyrin's user avatar
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8 votes
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Pressure at which graphite and diamond are in equilibrium

For each phase $i$ (graphite or diamond) you can show that $$\mathrm{d}\mu _i = V_i \mathrm{d}p - S_i \mathrm{d}T$$ or after integration $$ \mu_i = \mu_i^\circ+\int_{p^\circ}^{p} V_{i} dp $$ at ...
Buck Thorn's user avatar
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8 votes
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Gibbs Free Energy Notation

$G$ is the Gibbs free energy of a system. It is a conceptual quantity in the sense that there is no reference point that defines $G = 0$ for a substance (unlike entropy). Whenever you see a plot of $G$...
Karsten's user avatar
  • 40.7k
8 votes

How to estimate the temperature needed to overcome an activation energy barrier?

You could convert the rate constant($k$) to half-life($t_{1/2}$) which would give you an idea of the time scale required for the reaction to finish at a certain temperature. The equation to obtain ...
S R Maiti's user avatar
  • 5,685
8 votes

How to estimate the temperature needed to overcome an activation energy barrier?

From the Eyring equation, we can simply calculate the $k$ value for it. \begin{align} k &= \frac{k_\mathrm{b} T}{h}\exp\left(\frac{-\Delta G^\ddagger}{RT}\right)\\ k_\mathrm{b} &= \pu{1.38E-9 ...
M.L's user avatar
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7 votes
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Why is $\Delta G = -T \Delta S_{\mathrm{total}}$ valid only at constant pressure?

You have actually assumed in your equations that both pressure and temperature are constant. Firstly, we have that $$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} + \mathrm{d}S_\text{...
Linear Christmas's user avatar
7 votes

Gibbs Free Energy and Maximum Work

In order for a process to happen, it has to increase the combined entropy of the system in which it happens and of the surrounding (2nd law of thermodynamics). As we will see in a bit, the more work ...
Karsten's user avatar
  • 40.7k
7 votes

Is temperature double-counted in the Gibbs free energy equation?

One reason we write equations this way is because all of the parameters on the right-hand-side can be measured or computed independently. Consider an analogous (but more intuitive) equation, the ...
Buck Thorn's user avatar
  • 22.3k
7 votes
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Total Differential of Chemical Potential

$$dG=VdP-SdT-\sum{\mu_in_i}$$So, $$\frac{\partial G}{\partial P}=V$$and $$\frac{\partial G}{\partial n_i}=\mu_i$$So, $$\frac{\partial^2 G}{\partial n_i \partial P}=\frac{\partial^2 G}{\partial P\...
Chet Miller's user avatar
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6 votes
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How can a nonspontaneous reaction occur?

When you add energy from somewhere, let's say a battery, you are creating a greater amount of entropy there (in the battery or whatever). That's all there is to it. The Second Law is not violated. The ...
orthocresol's user avatar
  • 71.4k
6 votes

How to derive the Gibbs free energy for an ideal gas?

The fundamental relation for an Ideal Gas, in the entropy representation is $$S(U,V,N) = NS_0 + NR\ln\left[ \left( \frac{U}{U_0} \right)^c \left( \frac{V}{V_0} \right) ...
getafix's user avatar
  • 8,495
6 votes

How to derive the Gibbs free energy for an ideal gas?

Yes you are partly on the right track. Your final equation is usually written in a more familiar form as $$G=G^{\ce{o} }+RT\ln\left(\frac{p}{p^{\ce{o}}}\right)$$ and letting $p^{\ce{o}}$ equal to $...
porphyrin's user avatar
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6 votes
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Why are condensation reactions endergonic?

One of the most simple biosynthetic reactions that generates ‘oligomers’ from ‘monomers’ (between inverted commas because that description is not fully correct here) is the ester formation used to ...
Jan's user avatar
  • 68.2k
6 votes

Why can't a reaction go to completion?

Macroscopic view: $$\Delta G = \Delta G^{\varnothing} + RT\ln Q$$ As the reaction progresses, $Q$ gets larger so $\Delta G$ becomes more positive. At some point, $\Delta G = 0$ but this happens ...
Zhe's user avatar
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6 votes

Can we force reactions with positive delta G?

The change of the Free / Gibbs enthalpy $\Delta_{\text{R}}{}G$ consists of a contribution by reaction enthalphy $\Delta_{\text{R}}H$, and a contribution by reaction entropy, $\Delta_{\text{R}}S$: $$ \...
Buttonwood's user avatar
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6 votes

Why is my conclusion inconsistent with the van't Hoff equation?

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $\Delta G^\circ$ definitely depends on the temperature (and consequently on its ...
Karsten's user avatar
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6 votes
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How to explain disagreement between Le Châtelier's principle and the simplified Gibbs free energy equation?

The primary flaw in your reasoning is assuming that $K$ is proportional to $-\Delta G^\circ$, so that a reaction with $\Delta S^\circ >0$ and $\Delta G^\circ<0$ must have a larger $K$ at a ...
Andrew's user avatar
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6 votes

Are all exothermic reactions spontaneous?

No, it is not the case that all exothermic reactions are spontaneous. More generally, it doesn't even make sense to make a broad statement that a reaction is or is not spontaneous without also ...
theorist's user avatar
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6 votes
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Is there an algebraic form for the textbook reaction coordinate curves?

Here is a function that works: https://www.desmos.com/calculator/cqwgfj2nzt The general function would be $$ a \left( \frac{b}{1 - e^{5-x}} - \frac{c}{1 - e^{10-x}} \right)$$ I'm sure you could ...
Karsten's user avatar
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