Reverse process, enthalpy

FIGURE 6.23 The enthalpy change for a reverse process has the same value but the opposite sign of the enthalpy change for the forward process at the same temperature. 

If the dependence of the EMF on the temperature is known, then the reaction enthalpy AH of the reaction proceeding outside the cell can be found. Cells with a positive EMF, E, and a negative temperature quotient BE 1ST release heat in addition to electrical energy during reversible processes... 

R = Et, Pr), prepared independently, undergo unimolecular homolysis (effectively the reverse process) and the activation enthalpy for these processes were determined.838... 

Because most species will release energy when a proton is added, the proton addition enthalpy is negative. The proton affinity is the heat associated with the reverse process, the removal of the proton,... 

In summary, the foregoing examples show that for a given elementary reaction, the standard reaction enthalpy is derived from the difference between the enthalpies of activation of the forward and the reverse process. An identical conclusion is drawn for the entropic terms. If, in the cases of reactions 3.1 and 3.10, the rate constants k and k- are known as a function of temperature, those kinetic parameters may be determined by plotting In(k/T2) or In(k/T) versus l/T(k = k or k- ). This analysis is known as an Eyringplot, and the resulting activation enthalpies and entropies refer to the mean temperature of the experimental range. 

This definition may appear somewhat counterintuitive, because the word affinity suggests that we shouldbe referring to the reverse process. That is why Eea is often given as the negative of the enthalpy of reaction 4.8 at T = 0 [62], The definitions are, of course, equivalent and ensure that in most cases, electron affinities will have positive values. 

The van t Hoff plots for thermal denaturation of proteins are linear in the transition region, thus allowing the enthalpy change (AHm) of unfolding at the transition temperature (Tm) to be estimated. Because of the change in free energy in (AG) = 0 at Tm (reversible process), the entropy of unfolding (ASm) at the transition midpoint can be calculated from ... 

Temperature and enthalpy are not the only conditions that determine whether a change is favourable. Consider the process shown in Figure 7.5. A closed valve links two flasks together. The left flask contains an ideal gas. The right flask is evacuated. When the valve is opened, you expect the gas to diffuse into the evacuated flask until the pressure in both flasks is equal. You do not expect to see the reverse process—with all the gas molecules ending up in one of the flasks—unless work is done on the system. 

The molar reaction enthalpy ArH of the oxidation consists of work and heat. The second law of thermodynamics applied on reversible processes yields... 

The fall in temperature from 7 to the environmental temperature T0 releases an amount of enthalpy (A - h0), of which however only the exergy part can be used for the available and transformable energy to obtain useful work or products by means of reversible processes. We then define the energy availability A of a high temperature substance as in Eq. 10.14 ... 

Since the mixing is the reverse process of separation, the changes in enthalpy and entropy are usually negative and positive, respectively. Therefore the vector appears in the second quadrant on the thermodynamic compass. 

Unlike work and heat, the property changes of the system for step d can be computed, since they depend solely on the initial and final states, and these are known. The internal energy and enthalpy of an ideal gas are functions of temperature only. Therefore, A Ud and Atfd are zero, because the initial and final temperatures are both 27°C. The first law applies to irreversible as well as to reversible processes, and for step d it becomes... 

What happens in the reverse processes, when water vapor condenses to hq-uid water or liquid water freezes to ice The same amounts of energy are released in these exothermic processes as are absorbed in the endothermic processes of vaporization and melting. Thus, the molar enthalpy (heat) of condensation (A//gojjd) the molar enthalpy of vaporization have the same numerical value but opposite signs. Similarly, the molar enthalpy (heat) of solidification (A/Zg iid) and the molar enthalpy of fusion have the same numerical value but differ in sign. 

Recall from Section 12.1 that a true reversible process is an idealization it is a process in which the system proceeds with infinitesimal speed through a series of equilibrium states. The external pressure therefore, can never differ by more than an infinitesimal amount from the pressure, P, of the gas itself. The heat, work, energy, and enthalpy changes for ideal gases at constant volume (called isochoric processes) and at constant pressure (isobaric processes) have already been considered. This section examines isothermal (constant temperature) and adiabatic (q = 0) processes. 

About the same value can be calculated using Eq. (4.24) if Q - = 0, because the enthalpy change for a reversible process for 1 lb of water going from 100 psia and 100 F to 1000 psia is 2.70 Btu. Make the computation yourself. However, usually the enthalpy data for liquids other than water are missing, or not of sufficient accuracy to be valid, which forces an engineer to turn to the mechanical energy balance. 

Heat supplied is identified with enthalpy change because pressure is constant (for these reversible processes at constant P, q ev = = A//). 

The kinetics of reversible decompositions are often highly sensitive to reaction conditions [43]. For example, the values of and E, for the decomposition of CaCOj show unusually wide variations, owing to the sensitivity of reaction rate to the availability of COj [44,45]. The spread of apparent E values is considerable [46] and some values are close to the dissociation enthalpy [1]. However, Beruto and Searcy [47] concluded that, under high vacuum conditions, the constant rate of interface advance in large crystals was probably controlled by the dissociation step in the absence of a perceptible contribution from the reverse process. The decomposition activation energy (205 kJ mol ) was appreciably larger than the dissociation enthalpy (178 kJ mol ). This is probably the most precise kinetic measurement for the calcite decomposition [48]. 

For a reversible process, the dissociation enthalpy of a solid reactant may be determined from the variation of equilibrium decomposition pressure, pe, with temperature, for example, the dissociation of calcium carbonate (17) ... 

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