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4 <br /> Review of M902 Properties <br /> and the change in free energy is (using a finite difference designation rather than the <br /> differential or partial differential form): <br /> AF = AH - TAS - SAT [7j <br /> ~There H is the enthalpy (or heat energy) and S is the entropy. For our case, the change <br /> in enthalpy is very large and very negative. A large amount of heat is given off. This is <br /> negative because it leaves the system. The entropy change is positive and the <br /> temperature change is positive. This means that the free energy change of either <br /> equation [i] or [2), as written, is very large and very negative. This also means that the <br /> equilibrium constant K is very large and positive. <br /> To be able to use equation [7] to calculate En e-:.,ilibrium constant, It f5 customary to <br /> thin'K in terms of a reaction 1 here the tempera""!ire does not c;i=nge, i.e., the isothermal <br /> condition. Since in our m2nul3cture of the r,,a`e,i�l vrY con`.Col Che temperature, this is <br /> correct. Even in the normal use for the prod jct, the tempera's.:re changes are small <br /> (tens of degrees) compared to the changes t l?`. usually e�iec-i tide enthalpy and <br /> e,-)`.ropy. If vie elimin2t9 the lest term in equation [7) (since AT - 0), we h2`12 the more <br /> f��millar expression for free energy for %v,tch ste�dara L'aii.'esC'i t le free energy <br /> • iunctions have been determine-d over the yea,s. <br /> For the case of equation (I], N e have Una fa;loving free enerC:ea and enthal;)ies for the <br /> individual materials at 25°C. <br /> LHAF" <br /> l�callmole f;callmo'e <br /> NIg0 -142.92 -i 36 <br /> H202 I --14.85 I -28.78 <br /> M902 � -148.9 � -141.60- <br /> H20 <br /> 141.6?H2O -68.39 -60.657 <br /> Overall -29.49 -33.6 <br /> Although 1 hYve in the past calculated the entropy terms and calculated values for <br /> higher temperatures, I have not recalculated then for this r.v,mo. I believe my comment <br /> • in a previous memo that I did not calculate the entropy was taken to mean that 1 did not <br /> consider it. If one has the free energy values, there is no need to calculate the entropy <br /> terms unless one is Interested in ho,v those terns vary. <br />