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20

Yes it would, by a few percent. It may or may not be a goal worth pursuing, but there is more to it. Different reactions would be slowed down to a different extent. Tiny as they are, these discrepancies suffice to disrupt the delicate biochemical machinery of the living cell. No life form more advanced than bacteria is OK with that. Deuterated water in high ...


17

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 regarding how the world works, and our justification is that no experiment has ever shown the second law incorrect. In the framework of macroscopic thermodynamics, ...


16

No. The reason why a gas particle in a large volume has a large entropy is not because it has a lot of space to move around per se. A better explanation is that for a given energy, there are many accessible translational states (these states can be derived from the particle in a box model). If we assume that all of these translational states are equally ...


15

A process is thermodynamically reversible if it is essentially at equilibrium. Specifically, the system and its surroundings stay infinitesimally close to equilibrium with each other throughout a reversible process. Small changes in intensive variables of the system are perfectly balanced by changes in those variables in the surroundings. For example, $T_{\...


14

While it may seem that randomness always increases when a crystal is dissolved into a liquid phase, it does not have to be that way. Concerning sugar, the molecule has a large number of hydroxyl groups and is generally rather large when compared with the water molecule around it — much larger than your average sodium or chloride ions. Every hydroxyl will ...


12

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 movement tends to introduce randomness more than it tends to introduce order. Consider a deck of cards. Shuffle it. Is it perfectly sorted? No. Why? There are 10^67 ...


11

Without significant mixing, diffusion takes a long time to mix gases. Our understanding of entropy tells us that we will indeed finish with mixed layers, but that doesn't give us a time frame for that mixing, only an outcome. Given a slow but steady production of a dense gas, layers absolutely will form due to density differences. There are plenty of ...


11

Yes, the kinetic isotope effect is the main reason due to the differential lowering of the zero point energy in reactants and transition state, which has the effect of increasing, slightly, the activation energy. However, this effect is small, a few percent in a single reaction. The reason that deuteration has an effect overall is due to the fact that ...


10

For example, suppose you have a solid block of TNT. It explodes and releases much energy. $\Delta H$ is negative. Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased.


10

The most common way of measuring $\Delta S^\circ$ for a chemical reaction is probably by making a van't Hoff plot. You measure the equilibrium constant $K$ at different temperatures and plot $\ln K$ vs $T^{-1}$. The $y$-intercept = $R\Delta S^\circ$ and the slope = $-R\Delta H^\circ$. Another option is to measure $\Delta H^\circ$ by calorimetry and measure ...


9

Briefly, spontaneous processes tend to proceed from states of low probability to states of higher probability. The higher-probability states tend to be those that can be realized in many different ways. Entropy is a measure of the number of different ways a state with a particular energy can be realized. Specifically, $$S=k\ln W$$ where $k$ is Boltzmann's ...


9

The second Law of Thermodynamics states that the entropy of the universe always increases. $$\text{d}S > 0$$ In the case of adsorption the entropy of the system; the gas being adsorbed; decreases but the entropy of the surroundings;the rest of the gas and the surface (and everything else in the universe); increases and this outweighs the decrease in ...


9

NIST webbook does have a lot of data, though they are not in any kind of an API form as far as I know. http://webbook.nist.gov/chemistry/


9

"Reversible" is not binary. Both the forward and backward reactions always occur and the equilibrium system never has zero reactants or zero products. Thus, irreversible reactions are called this not because they cannot be reversed - they absolutely can - but because reversal is impractical. The equilibrium constant may be so skewed toward product that ...


9

All right, someone bearing the standard of thermodynamics will give you the equations shortly... From a layman to another, here goes my attempt at a simpler explanation. Entropy may be seen as the "disorderlyness" in some settings, but that is not a very useful way of seeing it. The metaphor is often used, but creates the wrong conclusions when looking ...


9

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 'particles' or quanta (say vibrational or rotational quanta in a molecule) can be placed among the various energy levels available. Thus at zero energy all the ...


8

Your definition of entropy is incorrect. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. $\mathrm{\delta S=\cfrac{\delta q_{rev}}{T}}$ (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously. We ...


8

There is a lot of uncertainty regarding the far future of our Universe, but it seems that chemistry as we know it will be gone long before the end. Both free and bound protons (and neutrons) are expected predicted to decay through at least one of several mechanisms, with a half-life somewhere between the range of $10^{35}$-$10^{200}$ yr (far shorter than the ...


8

You can't measure entropy directly, any more than you can measure interatomic distances. You measure other quantities -- for instance often you can measure energy gain/loss and temperature, and then you integrate $dS=dE/T$. How to explain it? One of the best expositions I know is The Second Law by Henry A. Bent. It is full of insightful examples, lays ...


8

NIST is the best place to turn for lots of data. However, more easily parsed, smaller datasets are available in a couple of other locations. The CHNOSz package in R has thermochemical data for a variety of species, mostly inorganic. Their database is referenced back to the chemical literature. See an answer I gave to an old question for an example of how ...


8

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 reaction $\ce{A <=> B}$. The reason that there is a finite and not zero back reaction rate constant, is that the activation barrier going B to A is not ...


8

The reason behind this is the hydrophobic effect. Everyone has seen it if they pour a spoonful of vegetable oil into a pot of water, e.g. to cook pasta. As long as nothing is disturbing the vegetable oil, it will collect itself together in one big bubble rather than form many small bubbles. Polar solvents will always be arranged in a way that positively ...


8

I consider watching any video a waste of time, so I'll be judging from your words alone. (Anyway, your question is essentially self-consistent, which is good.) Yes, entropy is a measure of disorder (sort of). No, $dS={\delta Q\over T}$ only for reversible processes. And no, entropy is not the Q-T ratio for two reasons. First, because the above formula deals ...


7

A simple explanation would be that the "o on top" denotes standard state and the one without the "o on top" denotes conditions that are not standard state. However, that may not allow you to "get it" so let's look at the equations. $\Delta G^{\circ} = -RT\ln K$ $\Delta G = \Delta G^{\circ} + RT \ln Q$ The first one allows us to find the Gibbs free ...


7

The entropy change between two thermodynamic equilibrium states of a system can definitely be directly measured experimentally. To do so, one needs to devise (dream up) a reversible path between the initial and final states. Any convenient reversible path will do, since the integral of dq/T is the same for all reversible paths. So you have to identify a ...


7

Consider two tanks of water. One hot, one cold. You open a valve between the two. The temperature between the two bleeds, and the hot tank loses entropy, and the cold tank gains entropy. However, entropy is also gained just by this process happening, as it would be impossible to force the 'cold' tank to retransfer the heat back to the 'hot' tank. Also, you ...


7

Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$. Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation $$\begin{align} Q_\mathrm{tr,...


7

The definition of entropy in 'Classical Thermodynamics' is $$\Delta S = \int\frac{\delta Q}{T}$$ The quantity on the left is the entropy change associated with a physical process. $\delta Q$ is an inexact differential, made exact by an integrating factor $\frac{1}{T}$ This notion was devolped by Kelvin and Cartheodory (I believe). This comes out of the ...


7

If you don't mind me asking, why would you expect the $\ce{Na}$ to have a higher entropy than the $\ce{NaCl}$? The entropy of a state is defined as,$$S=k_b\ln W$$where $W$ is the number of possible microstates which can make up that macrostate. In this case, we are interested in the number of possible ways that we can arrange $\ce{Na}$ into a lattice or the ...


6

If you write $\Delta G = \Delta H -T\Delta S$, you implicitly admit that temperature is constant and is the same for a non-isolated system and its environment. You basically assume that the system and the environment are in thermal equilibrium, at a given temperature $T$. Thermal equilibrium means that when system loses heat, environment absorbs it: this ...


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