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An interpation of the implicit derivative by the marginal rate of substiution makes this clear Marginal rate of substitution In economics, when the level set R(x, y) = 0 is an indifference curve for the quantities x and y consumed of two goods, the absolute value of the implicit derivative dy / dx is interpreted as the marginal rate of substitution of the ...

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Maybe, what they expect you to do is to write: $$-\left(\frac{\partial S}{\partial P}\right)_T=\frac{\partial ^2 G}{\partial P\partial T}$$From Eqn. 1, this gives, for an ideal gas $$-\left(\frac{\partial S}{\partial P}\right)_T=\frac{nR}{P}$$

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I would like to give an alternative derivation for theorist's answer. You can get the same answer with statistical mechanics using a lattice model. Let us discretize the container into $M_1$ respective $M_2$ cells. The volume of a cell is $V_{cell}$. Each gas particle fits exactly into one cell and can hop from cell to cell in a discrete manner. Since we are ...

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For an ideal gas at constant temperature $\Delta H=0$, so $\Delta G=-\Delta (TS)$, and, at constant temperature, $$\Delta G=-T\Delta S$$

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There's a simple way to solve this: We know that, for an ideal gas at constant T: $$\Delta S = -nR \ln \frac{p_f}{p_i}= nR \ln \frac{p_i}{p_f}$$ And since ideal gases ignore each other, we can calculate the entropy change for gas 1 and gas 2 separately, and simply sum the two. [At any given $T$, the chemical potential of an ideal gas is determined only by ...

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The thermodynamic data refers to formation of the given compounds from the elements in their standard states. For instance for $\ce{ ZnO(s)}$ it refers to the reaction $\ce{ Zn(s) +1/2O2(g) ->ZnO(s)} \tag{1}$ with each element or compound present at a pressure of 1 bar and (unless noted otherwise) a temperature of 298.15 K. A more negative value of $\... 1 You determined the$\Delta Hin going from products to reactants at constant temperature and volume. But, you are dealing with an ideal gas mixture, so the enthalpy of the product mixture is a function only of temperature, and not pressure. So the enthalpy change of the product mixture in going from the final pressure in the reactor to the initial ... 2 I think you can look at this entirely from the perspective of entropy (forget about the Gibbs free energy for a moment). You can divide the universe into two parts, the system and its surroundings. When a change happens in your system, and some heat is exchanged with the surroundings then the change in entropy of the surroundings is: $$\mathrm{\Delta S_{surr}... 0 The total differential of the internal energy (at constant composition and allowing only pV work) is usually written as$$dU = -pdV + TdS$$However the differential dU can be written in terms of other variables, for instance p and T, in which case:$$\begin{align} dU &= \left(\frac{\partial U}{\partial V} \right)_T dV + \left(\frac{\partial U}{\partial ... 1 We define a state function enthalpy as $$H=U+PV$$ Since we cannot measure the absolute value of enthalpy, but the change in it, we modify the equation to $$\Delta H= \Delta U +\Delta(PV)$$ Which is the correct relation. Now, if you assume pressure is constant, then you can take theP$out to get $$\Delta H=\Delta U+P\Delta V$$ Now$PV$work (or the ... 0 Enthalpy is the heat released during a chemical process under conditions of constant pressure. Therefore all the equations you have presented are correct. Expression two is the most specific definition of enthalpy. There's no reason P can't be a constant in the first equation (although it's appearance inside the bracket admittedly gives the impression that ... 1 The modern definition of thermodynamic stability is the state of maximum entropy. Some background information is necessary to make sense of this. I hope you will find the following helpful! Phenomenologically, thermodynamic stability is the absence of visible change. This is the 'original' definition, employed by experimentalists during the 18th and 19th ... 7 The full phrase should be thermodynamic stability with respect to ____, where the dash indicates a process, or a chemical reaction. A mixture of hydrogen and oxygen is thermodynamically unstable with respect to water formation. Similarly, a diamond is not forever (which may not please De Beers and ladies). It is thermodynamically unstable with respect to ... 0 The internal energy of a closed system is$dU=TdS-PdV $. In an isolated system$dU=0, dV=0$which would imply that$dS=0$, However, the entropy can increase due to chemical reaction or diffusion mixing different substances which were initially separate. Similarly$dG=-SdT+VdP$and if$T$and$P$are held constant this does not mean that$G$cannot change as ... 3 Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose) $$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$ And since$K_p$is related to$K_c$by the relation $$K_p = K_c(RT)^{\Delta n_\... 0 In less mathematical terms, the chemical potential of a component in a mixture at a specific temperature and pressure is the amount of free energy that can be ascribed to one mole of that component under the conditions found in the mixture at that T and p. Alternately it is how much the free energy of the system, if it were scaled to an infinitely large size,... 4 The error comes from reading the graph: the crossing of the axes corresponds to the point (2,2) and not (0,0) 0 The initial state of your system consists of 4 moles of \ce{Fe(s)} and 3 moles of water vapor at \pu{298 K} and \ce{1 bar} in separate containers. The final state of your system consists of 2 moles of \ce{Fe2O3(s)} and 3 moles of \ce{H2(g)} at \pu{298 K} and \pu{1 bar} in separate containers. The \Delta H^\circ and \Delta S^\circ correspond ... 2 When a system changes from state 1 to state 2, the change in entropy of the system is the same for all processes, because entropy is a state function. The difference between a reversible and an irreversible process is that, in the latter case, there is net entropy production for the universe, i.e., for the system+surroundings. Since the entropy change for ... 1 The way that you describe it is exactly correct. This is the essence of the Clausius Inequality. 2 I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample. This ... 3 I'm not going to do your research for you (at least not for now—perhaps if the mood takes me I'll do the Mathematica programming at some later point), but I will give you guidance that will help you find the answer: It's hard to find data on melting points at different pressures for a wide range of compounds (and you want to search through a wide range of ... 1 Gardeners often use a tennis ball for this purpose - it preserves a gap in the ice. 2 The equation$$\ce{2A <=> B + C}$$implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be$$\ce{(x + y) A <=> x B + y C}$$Here is an example that would be well-described by reaction (1):$$\ce{2 H2O <=> H3O+ + OH-}$$One water molecule cannot make a ... -1 You have stated that the reaction:$$\ce{2A <=> B + C}\tag{1}$$[...] can be thought the sum of the following reactions:$$\ce{A <=> B}\tag{2}\ce{A <=> C}\tag{3}$$However, this is not always the case. In particular, your assertion that the overall reaction (1) is a linear combination of reactions (2) and (3), is based upon the ... 1 There are oils sold for swimming pools to slow or stop evaporative cooling . That would be a problem for air breathers like mosquito larva. I don't know anything about it because I went with 1" thick Styrofoam panels. That is what I currently have on my pond ( 10' X 5 '), works very well .I cheat a little and have 150 watt aquarium heater in the 700 ... 0 A possible more natural solution allowing continuing oxygen exposure, use a layer of leaves resting on a frame (say wood branches). Supporting related comments can be found here, to quote: Leaves can be used to insulate plants from cold weather. To provide protection for plants, including container plantings, circle a plant with wire fencing. Then, stuff ... 2 Is the total entropy change of all isothermal processes 0? No. Only for reversible processes. However your equation is correct. For an irreversible process in which an ideal gas expands isothermally against a constant external pressure$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$becomes$$\Delta S_\mathrm{total}=\... 8 Everything you've written is correct (for a free expansion,$\Delta U = q = w = 0$), but it applies only if the system is an ideal gas, or has ideal-gas-like properties, namely that there are no attractive or repulsive interactions between the particles. It also requires that the system is closed (no matter can flow into or out of the system). Internal ... -1 To add to @Kexanone's answer above, the enthalpy of formation refers to how much heat is evolved or adsorbed from the environment in a reaction. So it's the total energy, and in an ideal gas it has the form: dft + translational + rotational + vibrational + PdV, where the PdV refers to the energy needed to make space for the gas (grand canonical ensemble; in ... 4 Imagine that you propose to climb Mount Everest. From your desk you can compute exactly how much energy you will need to reach the top of the mountain starting from base camp. That would simply be the gravitational energy difference between the two locations. With that information at hand you load up on supplies - dividing the gravitational potential energy ... 2 At low pressures the activity of a gas can usually be related in a straightforward way to its partial pressure as $$a=p/p^{\circ}$$ Since the activity of a substance present in multiple phases is the same in all of the phases, for a volatile compound it suffices to determine the partial pressure in the gas phase to know its activity in condensed phases. This ... 7 How is it possible, and what does it imply, when we say that the sum of two inexact differentials is an exact differential? In the example at hand it means that we divide the ways we can change the energy of the system into exactly two bins. The first we call heat, and all the other ones we call work. Heat is an energy transfer that occurs when two entities ... 9 There's not much to do, as you nearly spotted the difference by yourself:$U$is a state function, thus is path-independent. In other terms, one can estimate it at any moment only with an initial and a final value (and express it in terms of its variables only). But with$Q$and$W, it's different: these are path functions (or process functions) ... 4 The Gibbs free energy of a multicomponent mixture is a function not only of temperature and pressure, but also of the number of moles of the various species in the mixture: $$G=G(T, P, n_1, n_2. m_3. ...)$$So, $$dG=\frac{\partial G}{\partial T}dT+\frac{\partial G}{\partial P}dP+\frac{\partial G}{\partial n_1}dn_1+\frac{\partial G}{\partial n_2}dn_2+\frac{\... 2 The accepted answer is incorrect. It has nothing to do with reversibility. The underlining issue is that your dG is incomplete:$$dG = dH - TdS - SdT$$You are missing -SdT, so at constant pressure you don't end up with zero. As "ado sar" has commented in the accepted answer:$$dH = TdS + Vdp + \sum_i \mu_i \cdot dn_i$$Hence, you end up with: ... 7 G and H are like altitude. They are not absolute; rather their values are always relative to whatever you set as a reference. So they can be either positive or negative. There is no general constraint on the sign of their values. For this reason, we usually talk about \Delta G or \Delta H, since it's the changes that are meaningful; giving them ... 1 Yes, your expression is correct. It's nothing new though. Entropy of activation can be interpreted as part of the pre-exponential factor in the Eyring equation:$$k=\bigg(\frac{k_B T}{h}e^{\frac{\Delta^\ddagger S^\circ}{R}}\bigg)\cdot e^{-\frac{\Delta^\ddagger H^\circ}{RT}}$$0 Whether the lines are exactly concurrent depends on how the data were obtained. In the case of carbon oxidation, if we draw on measured thermodynamic data for oxidizing \ce{CO} to \ce{CO2} and also for oxidizing \ce{C} to \ce{CO2}, then we could calculate the "data" for oxidizing \ce{C} to \ce{CO} by doubling the first reaction and ... 1 The right way to do this is to understand the Van't Hoff equation. Barring that, I can provide some intuition, but it'll have to be a bit hand wavy. The way I think about this is that at higher temperatures, the intrinsic preference for reactants or products becomes less. That's because the thermal energy of the system is becoming more relevant than the ... 4 Because you are comparing apples with donuts. If I eat enough apples they will add more calories than that donut I gobbled for breakfast. To compare food calories properly you need to match serving sizes. Similarly, to compare reaction energies fairly you need to use a common basis for the extent of reaction. For oxidation of metals the reactivity of the ... 4 Old question, but still seems to attract some interest. In general, the total pressure of the system is not taken into considerations for corrections. The pressure at which you do the corrections is the partial pressure / concentration of the molecule. I recommend calculating the thermal correction for G at the temperature of interest and the default ... 0 The entropy change for ideal mixing is \Delta S=-R(x_A\ln{x_A}+x_B\ln{x_B}) 1 You are indeed right, I confirmed that it is the fault in the diagram itself. See the diagram by wikipedia: 1 Here are three examples of endothermic synthesis reactions. Heats of formation are given at 1 bar, 298.15 K. The first of these, the production of ozone, is naturally-occurring.$$ \begin{align} \ce{3/2 O2(g) &-> O3(g)} &\quad \Delta_\mathrm{f}H^\circ &= \pu{142 kJ mol^-1}\\ \ce{N2(g) + 1/2 O2(g) &-> N2O(g)} &\quad \Delta_\mathrm{... 4 Your definition of homogeneity is wrong Homogeneous systems are not, as you seem to assume, about the makeup or constituents of the system being uniform. Homogeneity is about the distribution of the components in the system being uniform. A solution of salt in water is homogeneous (if well mixed) despite having multiple types of molecule in it. If those ... 6 Any classification like chemical or a physical change, homogeneous or heterogeneous substance, will eventually fail at one point. From an analytical or physical chemistry's perspective, homogeneous substance is a substance, in a single phase, whose chemical composition is uniform. A heterogeneous substance has different composition from place to place like ... 5 General concept of heterogenity says the matter properties within the given region are location dependent ( within the given resolution and tolerance of the scenario ). The change of these properties can be the gradual ( like a concentration gradient ), abrupt ( like a phase boundary ) or combination of both. Practically speaking, matter is homogenous, if ... 0 It follows from the familiar result: $$\Delta H=q_p$$ By using the fact that $$C_p=\dfrac{q_p}{\Delta T}$$ We get $$\Delta H=C_p\Delta T$$ And take my note:\Delta H\$ is a state function so it doesn't matter how the process is carried out. Note, that the above answer is for an ideal gas, which the OP used in his question. Hope this helps. Ask anything if ...

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