Entropy change in reactions

One can interpret the log-uniform distribution through the Arrhenius law k — Aexp(—AG/fcT), where AG is the change of the Gibbs free energy inreaction (it includes both energetic and entropic terms AG = AH—TAS, where AH the enthalpy change and AS the entropy change in reaction, T the temperature). The log-uniform distribution of k corresponds to the uniform distribution of AG. 

Estimations. [39, 104] may be made from considerations of the equilibriuni constant K js and th,e rate coefficient for the reverse reaction is. The entropy change in reaction (—15) is —32 cal. deg . mole" and the enthalpy change will equal the dissociation energy )[C H2n—OOH] which has been estimated [39] to be 14 kcal.mole". Thus, -is = exp... 

Correction to put water on a 1 mol dm" basis raises it to 36 kJ mol", but the strong binding of several water molecules is clearly indicated. Note that the entropy change in reaction (15) is likely to be positive, since fewer water molecules would be bound around the larger ion. The activation energy for the decomposition of (CH3)2S S(CH3)2 in the reverse of reaction (15) has been found to be 57 kJ mol" [66]. The activation energy of the forward process is probably also appreciable, but has not been measured. 

This is an expression of Nernst s postulate which may be stated as the entropy change in a reaction at absolute zero is zero. The above relationships were established on the basis of measurements on reactions involving completely ordered crystalline substances only. Extending Nernst s result, Planck stated that the entropy, S0, of any perfectly ordered crystalline substance at absolute zero should be zero. This is the statement of the third law of thermodynamics. The third law, therefore, provides a means of calculating the absolute value of the entropy of a substance at any temperature. The statement of the third law is confined to pure crystalline solids simply because it has been observed that entropies of solutions and supercooled liquids do not approach a value of zero on being cooled. 

The metallothermic reduction of oxides is essentially a reaction involving only condensed phases. It follows therefore, that the entropy changes in these reactions are small and that the differences in the heats of formation of the pertinent compounds determine the feasibility of a given reaction. Among the metallic reductants, calcium forms the oxide whose heat of formation is the most negative. As a first approximation, calcium may be considered to be the most effective reducing agent for metal oxides. 

In much the same fashion as the AH° was tabulated, the standard molar entropies (S°) of elements and compounds are tabulated. This is the entropy associated with 1 mol of a substance in its standard state. Unlike the enthalpies, the entropies of elements are not zero. For a reaction, it is possible to calculate the standard entropy change in the same fashion as the enthalpies of reaction ... 

Q fiU What is the sign of the entropy change in each chemical reaction ... 

The enthalpy value of Eq. (3.23) is very small as might be expected if two Cd-N bonds in Cd(NH3) 2 are replaced by two Cd-N bonds in Cd(en). The favorable equilibrium constants for reactions [Eqs. (3.22) and (3.23)] are due to the positive entropy change. Note that in reaction, Eq. (3.23), two reactant molecules form three product molecules so chelation increases the net disorder (i.e., increase the degrees of freedom) of the system, which contributes a positive AS° change. In reaction Eq. (3.23), the AH is more negative but, again, it is the large, positive entropy that causes the chelation to be so favored. 

The variation of the association equilibrium constant, with reciprocal temperature is shown in Figure 6. These data yield a value of = -29.8 kcal mol for the enthalpy change in reaction (4), and AA = -26 cal mol K for the corresponding entropy change. As discussed previously, a combination of the... 

In this section we have so far emphasized only the AH° and AH° of the reaction components. This is because the entropy change in many reactions is small and can often be neglected in comparison to the enthalpy change. When S° s are of interest, they too can be estimated by Benson s additivity rules. In order to calculate 5° for a molecule, the group S° contributions are added together just as they are for AHf, but now a correction for the overall rotational symmetry (a) of the molecule must be added. The correction is — R In a, where a is the product... 

To determine the sign of AStotai = ASsys + ASsurr/ we need to calculate the values of ASsys and ASsurr. The entropy change in the system equals the standard entropy of reaction and can be calculated using the standard molar entropies in Table 17.1. To obtain ASsurr = —AH°/T, first calculate AH° for the reaction from standard enthalpies of formation (Section 8.10). 

RT log K, where K—pmJPp Pvi- The free energy of the formation of phosphine from hydrogen and solid phosphorus at 25° is —3296-0 cals. The entropy change in the reaction calculated from the free energy equation is —27-72 so that the entropy of phosphine at 25° is 52-4 units. J. C. Thomlinson compared the heats of formation of the trihydrides of the nitrogen-antimony family of elements. 

The association reactions of Si+" (2P) with acetylene and benzene have been measured by Glosik and coworkers96, who found that the rate coefficients for these reactions have strong negative-temperature dependencies. These observations were rationalized in terms of a negative entropy change in the reactions. 

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