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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 ...


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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 ...


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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 ...


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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. ...


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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 ...


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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.


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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 ...


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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^{\...


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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 ...


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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....


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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.


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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, ...


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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, ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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