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XXVII.

NOTES ON NEWTONIAN CHEMISTRY. BY REV. SAMUEL HAUGHTON, M.D., S.F.T.C.D.

[Read APRIL 11, 1892.]

NOTE II.-TERNARY COMPOUNDS.

(Water Type.)

In my first note on compounds of the hydrochloric acid type (HCl), I showed that the properties of the Halogen group, when combined with hydrogen, could be explained on Newtonian principles, viz.,

where

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is the coefficient of attraction, varying with the chemical nature of the compounds, m, m' the atomic weights, and r the distance of the bodies.

At all sensible' distances the coefficient of attraction (u) is constant ; but at insensible distances such as those contemplated in Chemical Science, μ ceases to be a constant, and varies with the chemical nature of the element, and varies also from other causes, as I shall demonstrate in this note.

By equating the centrifugal force to the attraction in any molecule the radius of whose orbit is unity, we found

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where is the angular velocity, m the reciprocal of w, and ẞ the atomic weight.

In sensible distances I include the smallest distances made known by the microscope.

R.I.A. PROC., SER. III., VOL. II.

2 K

Using the values found in Note I. we construct the following Table:

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This Table shows how enormously the coefficient of attraction of hydrogen exceeds those of the halogens.

In the case of hydrogen combined with one of the halogens, by equating the centrifugal force to the attraction, we obtain

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where ẞ is the atomic weight, and m the reciprocal of the angular velocity of the halogen molecule, and u' the coefficient of attraction between hydrogen and the halogen.

Using the values found in Note I. we construct the following Table, remembering that the values of m must be negative:

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1. Case of Water (H2O). In the case of water, two atoms of hydrogen are united to one atom of oxygen to form a molecule, and we must suppose that two molecules of hydrogen unite with one molecule of oxygen, with the result that the oxygen molecule is bisected, and that we have two molecules of water. The molecular volume of oxygen is the same as that of hydrogen, and (as is well known) the volumes of hydrogen and oxygen in becoming steam are reduced to two-thirds of the bulk of the original volumes.

The volume of the water (steam) molecule is the same as that of the hydrogen, fluorine, chlorine, bromine, iodine, and oxygen molecules, for

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a result strictly comparable with those already given for the molecular volumes of the elements named.

Before discussing the ternary compound of one atom of oxygen united with two atoms of hydrogen, let us consider the binary compound of one atom of oxygen with one atom of hydrogen. This binary compound is not known to exist separately, but is recognized in combination in complex bodies and is called hydroxyl. The reason why it is not formed separately is, that it is less stable than water,

H2O > HO,

1 Vide Appendix.

2 Hydroxyl. As a matter of fact, OH is not known in the free state, but is recognized by chemists as existent in numerous chemical compounds. All oxyacids are supposed to contain one or more OH groups, and the caustic alkalies, potash, soda, &c., contain OH also; while peroxide of hydrogen (H2O2) is regarded as an unstable compound of two OH groups which are tacitly assumed to be more or less opposed in properties.

This, as I shall show, is in entire accordance with the principles of Newtonian Chemistry.

An investigation of the properties of HO similar to that of the properties of HCl, in Note I., would give the rotations and stabilities of these two hydroxyls, viz.,

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and is thus prevented from making its appearance as a separate compound.

If we look back to the dynamical equations on which hydrochloric acid depends we have

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If we substitute in these equations for ẞ the atomic weight of oxygen we have the conditions necessary to account for the production of hydroxyl.

In equations (4) and (5) I have supposed the atoms to be infinitely hard, elastic, and without rotation, so that they cannot undergo any change of energy during collision, and that the whole change of energy takes place in the orbital changes of the molecules.

If we suppose this to be no longer the case, and that + €, denotes the atomic loss or gain of energy in forming one molecule of hydroxyl, we have, instead of equation (5),

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Eliminating

(5) bis.

between (4) and (5) bis, we find, after some re

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{2(8 + 1) + 1}m2 - 2ẞ2m - {B- B-B) = 0. (7) bis.

Substituting for ẞ the atomic weight of oxygen, we have

(346, + 1) m2 512m 3824 0.

If the atoms of oxygen and hydrogen have no rotation, and be infinitely hard and elastic, = 0, and the value of m becomes

m=-7.3625.

If we take (as in Note I.) the nearest whole number and make

m=-7,

we find,' as the most probable solution,

where

w'2 = 1.2564, a1 = 1.50, 1 = + 0·0042,

represents the atomic loss of energy, per atom of hydrogen, during the formation of hydroxyl, and w' represents the stability of hydroxyl.

1 See Note A, p. 430.

In the case of water (H2O) we have two molecules of hydrogen and one of oxygen converted into two molecules of water.

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There are four possible cases to be considered depending on the direction of rotation of the several molecules.

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