39

The classifications endothermic and exothermic refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$. If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you'...


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} ...


29

Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, $$\ce{H2O(l)->H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$ Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any ...


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 ...


19

The reaction coordinate is the progress of a reaction from reactants to products with various intermediates and transition states in between. It is an abstraction. It has no relation to time. Rather it is the progress of bond-forming and bond-breaking reaction steps. The free energy change of partially formed and partially broken bonds cannot be measured. ...


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 ...


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, ...


15

According to the principle of Microscopic Reversibility all elementary reactions, i.e. those that proceed in a single step, are reversible. And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your ...


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 ...


13

Hammett wanted to find a way to quantify the effects electron-withdrawing (EWG) and - donating (EDG) groups have on the transition state or intermediate during the course of a reaction. Initially he took the $pK_{\mathrm{a}}$-values of benzoic acids carrying the respective functional group in para or meta position (ortho acids and aliphatic acids weren't ...


13

Pressure-Volume work is a type of mechanical work . There are also non-mechanical forms of work where in pressure and volume terms are not involved . The most famous example is electrical work . The work done to move electrical charges ( like motion of electrons in a conductor) comes under this category .


13

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 ...


13

$\Delta G^\circ_m$ 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 solid and that for the product is the solute in a $\pu{1 molal}$ ideal solution, all at $p^\circ =\pu{ 1 bar}$). If $\Delta G^\circ_m>0$ you cannot ...


12

There are two considerations about why water evaporates but only one of them has to do with the Gibbs free energy of vaporisation. As the first answer correctly states, there is an equilibrium (heavily biased in favour of the liquid not the vapour) at room temperature. In equilibrium (which is what the $\Delta G$ measures) there is enough thermal energy to ...


11

Gibbs free energy is a measure of how much "potential" a reaction has left to do a net "something." So if the free energy is zero, then the reaction is at equilibrium, an no more work can be done. It may be easier to see this using an alternative form of the the Gibbs free energy, such as $\Delta G = -T\Delta S$.


11

Equilibrium is also the state in which the forward and reverse reactions are proceeding at equal rate. Consider a hypothetical reaction $$\ce{aA + bB} \overset{k_1}{\underset{k_{-1}}{\ce{<=>}}}\ce{ cC + dD}$$ If we assume that the forward and reverse processes are elementary reactions (a big assumption), then: $$rate_1=k_1 [\ce{A}]^a [\ce{B}]^b$$ $$...


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

An example program is GAMESS. It is free. But be warned: the calculation of BDEs is an extremely difficult thing to do. I am presuming that you are a beginner, so you have to be careful for issues surrounding which BDE: \begin{align} \ce{A-B &-> A+ + B-} \tag{ionic}\\ \ce{A-B &-> A. + .B} \tag{homolytic} \end{align} Most electronic ...


9

Thanks for the edit - I see now what you meant. The short answer for why we "need" Gibbs energy is that the internal energy for a spontaneous process at constant temperature $T$ and pressure $p$ does not necessarily decrease. Internal energy is the thermodynamic potential (wiki link) for constant $S$ and $V$ - this makes sense if you think about it: If the ...


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

$\Delta G = -RT \ln K$ Which means there for any given process an equilibrium constant can be deduced. Given the products are 'gaseous' they will escape the system, and further drive the equilibrium. In this case, $K = 0.05$ Applied pragmatically, we define a system as 'near the puddle'. When the water evaporates, it can either condense again, or move ...


9

The association of the Gibbs free energy with “additional”, or “non-expansion” work, is simply a mathematical result of its definition: $G = H - TS = U + pV - TS$. From the First Law, for a closed system, we have $\mathrm{d}U = \delta q + \delta w$. For a reversible change, the Second Law tells us that $\mathrm{d}S = \delta q_\text{rev}/T$; so $\delta q_\...


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

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

The answer is both. The formula for the equilibrium constant was first observed experimentally (that is, it's empirical) for many many reactions. Then it was justified theoretically in terms of what is now called the Gibbs energy. The reason the formula is multiplicative is that (very roughly) the probability of finding a system in a certain state is ...


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

First of all, what exactly is spontaneous reaction? In very easy language, reaction that occurs in a given set of conditions without intervention is called spontaneous reaction. Now let us consider melting of ice example, Take a ice on a plate and leave it for half an hour(conditions). After half an hour we will notice water instead of ice. This ...


8

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 at this from the perspective of Hess's Law. How so? Well, if we add two reactions, no matter which ones, there's one certainty we keep in our souls: the new ...


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 ...


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