«Limitations to the Law of Chemical Equilibrium.»—Quite in agreement with the interpretation of the law of chemical equilibrium from the view-point of the kinetic theory, it is found that, in its applications, one must take into consideration the possibility, that molecular attractions (p. 27) or other important forces of attraction or repulsion (‹e.g.› electrical) of a more than ‹negligible› magnitude exist between the molecules of the components in a reversible reaction. If such forces are involved, suitable allowance must be made for them, so that the mathematical formulation of the law may express the facts of observation.[174] For these reasons, the law, in its simplest terms, which, alone, can be considered here, holds for the relations ‹obtaining in dilute systems›[175] (dilute solutions or gases that are not too strongly compressed), and in ‹systems involving only nonelectrolytes or only weak ionogens›, more generally than for the relations in concentrated solutions (or strongly compressed gases) or in systems, in which electrically charged particles (ions) are present in large proportions (see below, p. 108). «The Factors of the Law of Chemical Equilibrium.»—Inspection of the mathematical expression of the law of chemical equilibrium (p. 91) shows that there are two significant kinds of factors in it: first, the equilibrium ‹constant›, whose value depends only on the nature of the substances involved and on the temperature.[176] In [p096] the second place, we have concentration factors, which, to a large extent,[177] may be varied at will. The following experiments may be used to illustrate the significance of the two classes of factors: Phosphorus pentabromide is partially decomposed by heat into the tribromide and bromine (a case of gaseous dissociation): PBr_{5} ⇄ PBr_{3} + Br_{2}. Phosphorus trichlordibromide is decomposed more or less, in a similar fashion, into the components phosphorus trichloride and bromine, according to the equation PCl_{3}Br_{2} ⇄ PCl_{3} + Br_{2}. For the condition of equilibrium in the two cases we have [PBr_{3}] × [Br_{2}]′ / [PBr_{5}] = ‹k›_{1} and [PCl_{3}] × [Br_{2}]″ / [PCl_{3}Br_{2}] = ‹k›_{2}. EXP. Two tubes containing equivalent quantities of the two bromides are placed side by side in warm water.[178] The tube containing the trichlordibromide is found to be much more intensely colored by free bromine than that containing the pentabromide.