![]() The use of two or more inconsistent links to the same elemental reference form in a thermodynamic database will result in an inconsistency in the database. In evaluating a reported thermodynamic datum, one should identify the links to the chemical elements, a process which can be time-consuming and which may lead to a dead end (an incomplete link). Some or all of its associated data may also be key data. Such a link differs from a bare “key” or “reference” datum in that it requires additional information. A valid link consists of two parts: (a) the path of reactions and corrections and (b) the associated data, which are key data. This concept involves a more » documented understanding of all reactions and calculations leading to values for a formation property (standard Gibbs energy or enthalpy). We propose a formal concept of “links” to the elemental reference forms. Thermochemical convention is to define the standard Gibbs energy and the standard enthalpy of an individual chemical species in terms of formation from reference forms of the constituent chemical elements. A consistent set of primary key data (standard Gibbs energies, standard enthalpies, and standard entropies for key chemical species) for 298.15 K and 1 bar pressure is essential. Such data pertain to elemental reference forms and key, stoichiometrically simple chemical species including metal oxides, CO2, water, and aqueous species such as Na+ and Cl. One problem pertains to the use of inconsistent primary key reference data. However, problems with thermodynamic data appear to be persistent. The past several decades have seen the development of thermodynamic databases in digital form designed to support computer calculations. The last century has seen the development of a large body of thermodynamic data and a number of major compilations. If ∆S is negative, then the negative signs (from the subtraction and the sign of ∆S) will cancel out, and so as T∆S gets bigger, ∆G will get more positive.Chemical thermodynamic data remain a keystone for geochemical modeling and reactive transport simulation as applied to an increasing number of applications in the earth sciences, as well as applications in other areas including metallurgy, material science, and industrial process design. T is always positive, so if ∆S is positive then a bigger T∆S will make ∆G more negative (since we subtract T∆S). As T increases, the T∆S component gets bigger. ∆H is still positive and ∆S is still whatever sign you figured out above. Since ∆H and ∆S don't change significantly with temperature (given in the question), we can assume that they keep the same signs and values: i.e. If ∆G is negative (from the question), is the reaction spontaneous or non-spontaneous?Ģ) Let's use ∆G = ∆H - T∆S again. From these values, we can know for certain whether ∆S is positive or negative (hint: remember that we are subtracting ∆G!).ġ) Knowing the sign of ∆G is enough to say whether the reaction is spontaneous or not under these conditions. Temperature is always positive (in Kelvin). We know (from the question) that ∆G is negative and that ∆H is positive. This looks like a homework question, so I'll give you some hints to get you on the riht path rather than answering directly.ģ) We know that ∆G = ∆H - T∆S.
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