# Tag Info

0

Although I am conversant with this aspect of thermochemistry, it is not my area of expertise. Nonetheless, your question as to why branching leads to a lower heat of combustion in a set of alkane constitutional isomers is one that has intrigued me for sometime. So much so that, after asking former peers steeped in physical organic chemistry, they were at a ...

1

Branched-chain alkanes have lower values of ΔcH⊖ than straight-chain alkanes of the same number of carbon atoms, and so can be seen to be somewhat more stable. https://en.wikipedia.org/wiki/Alkane#Branched_alkanes The heats of formation are given in https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation and go from -40.0 kcal/mol for a to -41.8 for b ...

1

First, not all gases can be liquefied at room temperature by increasing pressure. If the gas is above the critical temperature, it cannot be liquefied by any increase in pressure; it becomes a supercritical fluid. Supercritical fluids have some of the properties of a gas (e.g. diffusing through fine openings), ans some of liquids (e.g. dissolving solids and ...

-2

As one increases the pressure, the volume occupied by the gas decreases. Like you said, the temperature should increase with increasing pressure. Equivalently you could say that the thermal energy of the molecules of the gas has increased which manifests itself in increased collisions and vibrations of the constituent molecules. The decrement in volume ...

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 ...

0

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 ...

0

Careful, if you have samples that contain transition metals and/or their salts, in the presence of ammonia fumes, air (a source of oxygen), water (or even moisture) and an electrolyte (any soluble salts), you may have some electrochemistry afoot as well. In particular, for example, is the leaching of copper ores (removing the copper ions) with NH3/air with a ...

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 ...

7

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 ...

0

If the container is open, as shown below, then you have a total mess. Obviously eventually all the ammonia and water will evaporate into the room eventually. There is absolutely no way to tell how much ammonia gas would be with the volume of the container at any given time, and the ammonia concentration within the container wouldn't ever be homogeneous. ...

3

There exist many different potential energies. Maybe you would prefer an analogy. For example let's consider an object on a table (a ball, a book or a piece of chalk). It has a potential energy which is not chemical for the moment. It is not visible. It is a hidden energy. It can only be shown if you give the object an opportunity to fall down and to go to a ...

1

It is necessary to differentiate between the change in system entropy and surrounding entropy. Rusting involves a decrease in system entropy, but causes an even greater increase in the entropy of the surroundings. This is because it is an exothermic reaction, releasing heat energy to the surrounding molecules, increasing their disorder. Overall, when you ...

1

As already noted by Maurice, yes you did liquify a gas under pressure. In fact, the scope of application is so wide that this is equally known as Linde cycle. It is highly possible that you have such an engine at home, either as fridge, or freezer to cool stuff, or as heat pump to warm a home. Regarding propane: equally yes. After banning Freon and other ...

0

This is a fun question. Given that you are looking for kJ/g rather than kJ/mol, it seemed most fruitful to explore the possible compounds that can be formed from the lightest-weight elements. So: I was unable to find any compounds to beat those in the lists provided by Jonathan and Matthew for exothermic heats of formation. But, considering compounds ...

3

In thermodynamics, the basis for a definition of temperature is provided by the $0^{\text{th}}$ Law: two bodies independently in thermal equilibrium with a third body are in thermal equilibrium with one another. Thermal equilibrium allows the definition of temperature: two bodies in thermal equilibrium are said to be at the same "temperature". The $0^{\... 7 Temperature is related to kinetic energy, but it can't be simply equated to the average kinetic energy of the system. As I wrote in response to another answer, different systems can have different average kinetic energies/particle, but the same temperature. E.g., at the same temperature the avg. kinetic/energy particle of a diatomic gas is greater than ... -4 Temperature is the average kinetic energy of the particles making up a system. That's it, and it is correct. Any other definition, and there are many of this page, are either equivalent or incorrect. What's the problem? 0 First, I present an example, which provides a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:$\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$Reference: one of many ... 11 Temperature vs kinetic energy [OP:] I've read at many places that temperature is the average kinetic energy of particles present in an object. Temperature has to do with the average kinetic energy of particles, but to say the two concepts are the same is incorrect. What is correct is that if the particles in two mono-atomic gas samples have the same ... 11 Heat is the transfer of energy to or from the body in forms other than matter flow or work (organized energy transfer, such as pushing). Temperature is only a well-defined property for a collective body (you wouldn't be able to tell me the temperature of a single atom, for example). Like you said, it's the property of matter describing the amount of kinetic ... 3 In a thought experiment, you can substitute the electrochemical half reaction by three steps: Removing atoms from the electrode into the gas phase (sublimation) Removing electrons from the atoms (ionization) Solvating the resulting ions If you have quantitative descriptions of these processes, you get a quantitative description of what is actually going on.... 1 Welcome to the Chemistry Stack Exchange! In general, heat transferred to an ideal gas is not considered as work that was done on the system. I suspect you're referring to a process in which the internal energy doesn't change:$\Delta{U}=0$. As$\Delta{U} = q + w$(where q is heat and w is work), in the case where$\Delta{U}=0$you get$q=-w$. This holds for ... 0 Here are two common ways of measuring the entropy change in a reaction: Measure the equilibrium constant$K$at multiple temperatures. This gives you the Gibbs energy (via$\Delta_r G^\circ = - R T \ln{K}$) and, via the van't Hoff relationship, the enthalpy. You can calculate the entropy from those two. Use microcalorimetry in a titrating mode. You get the ... 0 As an interesting bit of history, Boltzmann was the first one to describe entropy as a "measure of disorder" of a system. It's worth noting that he didn't know this was an oversimplification. In reality, entropy is best described as a measure of the number of ways that energy can be distributed in energy levels between or within particles. From this, it's ... 1 Don't know if these are quite in the spirit of what you are looking for since, with the exception of the first reaction, the entropy change alone is not sufficient to drive them (i.e., at room temp., their equilibria lie far to the left) (and, for ozone formation, the entropy change is actually unfavorable). Furthermore, ozone formation and photosynthesis ... 0 Here is a rather simple structure explanation presented on this site. To quote, where 'pieces' refer to an organic molecular with a non-symmetric branching structure: You’ll notice something – the simpler the pieces are, the easier they are to stack together, which provides a tighter fit with fewer spaces. Here, by putting a kink in the block, we make ... 0 Hyperconjugation is an electronic effect which is similar to resonance (one of the most fundamental concepts of organic chemistry). Here's how it works:$\ce{H3C-CH2+<->H+ H2C=CH2}$You can write the above structure three times for each hydrogen replaced. Note that only$α-H$'s (on the adjacent$sp3$carbon atom) can be replaced. Now, answering ... 0 One endothermic low-temperature reaction that gets a lot if use, seasonally, is salting of icy or snowy surfaces. The sodium chloride combines with the ice to form a liquid solution. Sodium chloride has little heat of solution either way with liquid water, but here the solid ice has to be converted to the liquid. Therefore we have a reaction that's ... 2 Assuming that by air bubble you're referring to the pocket of air now sealed in the bottle above the water surface: Immediately after sealing the bottle, the air pocket is not at all compressed: it's at the same pressure as the surrounding environment (the air outside the bottle). However, because you have a closed container, some water will evaporate, ... 3 They're for the entire bond (so for a triple bond, all 3). From Wikipedia, it can be defined as The standard enthalpy change when [a bond, be it single, double, or triple] is cleaved by homolysis to give fragments A and B, which are usually radical species So you get two fragments as your final state. 2 From Fundamentals of Thermodynamics by Borgnakke and Sonntag [1, p. 94]: Further consideration of a$P-V$diagram, such as Fig. 4.6, leads to another important conclusion. It is possible to go from state 1 to state 2 along many different quasi-equilibrium paths, such as A, B, or C. Since the area under each curve represents the work for each process, ... 5 For any process, the work done by the gas on the surroundings is$\int{P_\mathrm{ext}\,\mathrm dV},$where$P_\mathrm{ext}$is the force per unit area the piston exerts on the gas at the piston face, and also the force per unit area the gas exerts on the piston at the piston face. For a reversible process,$P_\mathrm{ext}$is also equal to$P,$the pressure ... 3 Because$\Delta H$is temperature-, pressure-, and concentration-dependent, it's certainly theoretically possible for$\Delta H$for a chemical reaction to change sign in response to changes in any of those intensive properties. I don't have an example of such a chemical reaction handy. However, there are direct experimental measurements showing ... 3 Water loses kinetic energy when it detaches from the surface of the ice Water molecule in the solid phase have more hydrogen bonds than in the liquid phase. It takes energy to break these bonds, so only molecules with sufficient kinetic energy will detach (maybe because they just collided with a water molecule that transferred some kinetic energy to it). So ... 0 A possible example (which is real and reflective of a class of compounds, energetics) is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like$\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously, ... 4 An exothermic reaction is a reaction for which the overall standard enthalpy change$\Delta H^\circ$is negative. An endothermic reaction is a reaction for which the overall standard enthalpy change$\Delta H^\circ$is positive. Clearly, it is impossible that$\Delta H^\circ$is simultaneously negative and positive. 1 Energy is needed for separating positive from negative ions in the dissolution process. This energy is taken in the surrounding water. Water is loosing energy in the dissolution process. That is why the temperature of the water decreases. There is nothing special in using$NH_4Cl$. The same phenomena happens when dissolving a salt like$NaCl$or any other ... 2 An exothermic reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released into the surroundings, resulting in an overall negative quantity for the heat of reaction (-ΔE ). An endothermic reaction occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system ... 0 At equilibrium, the forward and reverse rates have to be equal. For mass action kinetics, this tells you that the reverse rate constant must be equal to the forward rate constant divided by the equilibrium constant. 1 In general I would say 'no', thermodynamics only tells us about starting and ending points, thus you know the$\Delta G^\text{o}$for the reaction but nothing about its actual mechanism, (since time does not come into thermodynamics). As$K_\mathrm{eq}$is a ratio of rate constants these can take on many different values and still have the same ratio. To ... 2 From the abstract of the article in question: The results suggest that the pressure stability of a protein in solution is not directly affected by the presence of these proposed piezolytes, and so they cannot be granted this distinction. So the answer is "no", the supposed piezolytes do not appear to have the purported function. They do stabilize ... 3 There are a number of possible explanations for the entropy difference, which can be computed based on standard free energies and enthalpies of formation as$\pu{3.0 J/molK}$. First, it could be that the entropy change is so small ($TΔS^∘<\pu{1 kJ}$, compare this to$ΔH^∘=\pu{−393.5kJ/mol}\$) and measurement too imprecise for this difference to be ...

Top 50 recent answers are included