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电化学专业英语翻译
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Note than reactions 6,7,and 15 involve fractional stoichiometric coefficients on the left-hand sides:this is becaude we wanted to define conventional enthalpies of formation (etc.) of one mole of each of the respective products.However,if we are not concerned about the conventional thermodynamic quantities of formation,we can get rid of fractional coefficients by multiplyung throughout by the appropriate factor.For example,reaction 6 could be doubled,whereupon becomes △G,△H=2△Hf ,and △S=2△Sf,and the right-hand sides of Eqs. 13 and 14 must be squared so that the new equilibrium constang K=K*K=1.23*10^83 bar^-3 ,Thus,whenever we give a numerical value for an equilibrium constant or an associated thermodynamic quantity,we must make clear how we chose to define the equilibrium.the concentrations we calculate from an equilibrium constant will,ofcourse,be the same,no matter how it was defined.Sometimes,as in Eq. 14,the units given for K will imply the definition,but in certain cases such as reaction 15 K is dimensionless. 电化学专业英语翻译，谢谢！！ |
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