Past exam of the medicine course of the University of Oxford / 1864 / First examination / Chemistry / 8

Original question.

Give the ratio which equal volumes of the following gases bear to the volumes of the elements of which they consist :—Hydrochloric Acid, Water, Protoxide of Nitrogen, Binoxide of Nitrogen, Ammonia.

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- Solution 8

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At the same temperature and pressure, one volume of HCl corresponds to $\frac{1}{2}$ volume of hydrogen and $\frac{1}{2}$ volume of chlorine; one of steam to one of hydrogen and $\frac{1}{2}$ of oxygen; one of $N_{2} O$ to one of nitrogen and $\frac{1}{2}$ of oxygen; one of NO to $\frac{1}{2}$ of each element; and one of ammonia to $\frac{1}{2}$ of nitrogen and $\frac{3}{2}$ of hydrogen. All elemental volumes here refer to their ordinary diatomic gases.

The simplest way to avoid an erroneous factor of two is to balance each chemical reaction using

H_{2}

C l_{2}

N_{2}

and

O_{2}

. Under Avogadro's law, their stoichiometric coefficients are also gas volume ratios, provided all gases are compared at the same temperature and pressure and ideal gas behaviour is a good approximation.

For gaseous hydrogen chloride, $H_{2} + C l_{2} \to 2HCl$ . Thus $V_{HCl} : V_{H_{2}} : V_{C l_{2}} = 2 : 1 : 1$ . There is no change in total gas volume in this idealized combination.
For water vapor, $2 H_{2} + O_{2} \to 2 H_{2} O (g)$ . Thus $V_{H_{2} O (g)} : V_{H_{2}} : V_{O_{2}} = 2 : 2 : 1$ : three reactant volumes give two product volumes.
“Protoxide of nitrogen” is nitrous oxide, $N_{2} O$ . The composition equation $2 N_{2} + O_{2} \to 2 N_{2} O$ gives $V_{N_{2} O} : V_{N_{2}} : V_{O_{2}} = 2 : 2 : 1$ . This equation expresses composition; it does not assert that direct mixing gives an easy preparative reaction.
“Binoxide of nitrogen” in the historical equivalent nomenclature is nitric oxide, modern $NO$ , rather than modern nitrogen dioxide. The equation $N_{2} + O_{2} \to 2NO$ gives $V_{NO} : V_{N_{2}} : V_{O_{2}} = 2 : 1 : 1$ . Older $N O_{2}$ notation with oxygen equivalent 8 describes the same oxygen-to-nitrogen mass ratio as modern $NO$ with oxygen atomic weight 16; reading its subscript by modern rules changes the substance.^[ref]
For Ammonia, $N_{2} + 3 H_{2} \to 2N H_{3}$ gives $V_{N H_{3}} : V_{N_{2}} : V_{H_{2}} = 2 : 1 : 3$ : four reactant volumes correspond to two product volumes.

If the common chosen volume of each compound is

V

, the corresponding ordered elemental volumes are therefore

(V /2, V /2)

(V, V /2)

(V, V /2)

(V /2, V /2)

and

(V /2, 3 V /2)

. The ratios concern molecular counts and hypothetical free elemental gases, not separate compartments of elemental gas still present inside a compound.

“Hydrochloric acid” must here mean gaseous hydrogen chloride, rather than its aqueous chemical solution; water must remain as steam. If water condenses or Ammonia dissolves in the collecting liquid, measured volumes no longer show these simple ratios. For an experimental comparison, dry-gas and hydrostatic pressure corrections are those discussed in 3.

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Past exam of the medicine course of the University of Oxford / 1864 / First examination / Chemistry / 8

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