A solution between these two would be more probable than either, and we may find it by means of the following Table :

XXVIII.

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

[Read APRIL 11, 1892.]

NOTE III. QUATERNARY COMPOUNDS.

(Ammonia Type.)

1. Ammonia. We have now reached the third of Mendelejeff's four typical compounds, viz. that in which a triad molecule combines with three hydrogen molecules, of which the most perfect example is ammonia.

We have seen that the monad molecules combine together in binary groups, like hydrochloric acid, and in doing so, satisfy the two most important conditions for catastrophe in a planetary system, viz.:

1o. The motions of the combining molecules in the same plane are in opposite directions.

2o. The periodic times of the combining molecules have a small common multiple.

In the case of the dyad molecule combining with two monad molecules, the same conditions for catastrophe are fulfilled; and in addition it was found that the molecules could not combine unless assisted from without by electricity or heat appearing in the dynamical equations in the form of a bond fide addition to the sum of the areas (moments of momenta).

The volume of the nitrogen molecule is the same as that of the elements already discussed, for

The molecular volume of ammonia is also very nearly the same as that of nitrogen, for

In the present case, of which ammonia is the type, we have a molecule of a triad element combining with three molecules of hydrogen, and we shall show that similar conclusions hold. There are three combinations of nitrogen with hydrogen whose properties can be calculated on Newtonian principles; and although one only of the three is known in a separate form, the existence of the others in a combination in higher compounds is probable, and their non-appearance in a separate form can be explained by the greater facilities which exist for the production of ammonia, which precludes them from appearing.

These three compounds are

(a). NH.

(b). NH2.

(c). NH, ammonia.

(a). Nitride of Hydrogen in which nitrogen behaves as a monad (NH). The general equations belonging to this case are-

=

where B 14 in the case of nitrogen.

We therefore have to deal with the case of three equations containing five unknowns,

where

w'/m/a1/μ'/€1,

w2 represents the stability of NH.

m the periodic time of the nitrogen molecule.

a1 the addition made from without (if any) to the equation of areas per atom of hydrogen employed.

the coefficient of attraction of nitrogen for hydrogen in NH. the loss of atomic energy (if any) per atom of hydrogen employed.

We know also that w' is greater than unity, and that μ', a1, and e are [ositive.

As a first approximation to the solution of the three equations,

but since w' is greater than unity, we must choose the negative value of '; and since it follows that m must be negative when w' is negative, for

the molecules of nitrogen and hydrogen must, therefore, revolve in opposite directions, thus fulfilling the first condition for a catastrophe. The second condition for a catastrophe is that m should be, if possible, a whole number.

Our concern is with the negative root, which, however, does not fulfil the second condition essential to a catastrophe. Let us then assume m = - - 6 the nearest integer, and find values of a, and ε to suit our supposition.

which, being less than unity, must be rejected.

The second case is impossible, because & must be positive; and the first case may be adopted, because it gives

2€1 E

=

0.2637 per cent.,

which may be easily admitted.

From the value m1 = − 6 we find the attraction of nitrogen for hydrogen in NH to be

μ' = 31746.

1 This solution is improbable, because it throws the whole strain of the transformation upon e, the loss of energy incurred by the atoms, independent of the orbital motion, and it is easy to see that this loss of atomic energy is much too great to be admitted.

In fact, the total energy of the two molecules of nitrogen and hydrogen before the shock is

This makes the percentage of atomic loss to orbital energy to be 19.454 per cent., or nearly one-fifth, which would be enormously greater than anything known among our planetary experiences.

2 This solution supposes the atoms of nitrogen and hydrogen to be without rotation, and infinitely hard and elastic, which is improbable.