Page:Outlines of Physical Chemistry - 1899.djvu/250

 232 OUTLINES OF PHYSICAL CHEMISTEY

must, therefore, correspond to the difference of the initial states, that is, to the complete displacement of one acid by the other. Thomson has found that

[Na a S0 4 aq, KJ$ 2 6 aq]* = -3-504 Cal. [Na 2 N 2 6 aq, H 2 S0 4 aq] = +0-576 Cal.

By the displacement of the sulphuric acid by the nitric acid, therefore, 4*080 Cal. are absorbed, that is, 2*040 Cal. per equivalent. The difference of the heats, of neutrali- sation directly determined for an equivalent of sulphate and an equivalent of nitrate of sodium is 2*072 Cal.

Let us recall that the action of an acid on a salt of the same acid often gives a considerable thermal effect. We know also by the law of thermoneutrality that on mixing two neutral salt solutions generally no appreciable effect is produced ; and we may further add that the same remark applies to two dilute solutions of different acids.

We can now approach the main point of the subject and show how thermo-chemistry gives us a solution of the problem under discussion.

If we allow a dilute solution of one equivalent of nitric acid (a) to react on a dilute solution of one equivalent of sodium sulphate (b), stationary equilibrium will be esta- blished when a quantity x of the acid has reacted on the salt to form free sulphuric acid (a x ) and sodium nitrate (b\). We then have as before :

k(l-xy = k,x\ and


 * ?L — x%

The observed thermal effect will be due to two distinct causes :

1. The transformation of x sulphate into x nitrate ;

2. The action of x free sulphuric acid on the remaining

1 a* is a fractional number.

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