Question #ebf3f

1 Answer
Oct 6, 2016

Answer:

It is a fact (i) that most elemental gases are DIATOMIC; it is a fact (ii) that mass is always conserved in a chemical reaction.

Explanation:

Most of the elemental gases, #O_2#, #N_2#, #F_2#, #Cl_2# are diatomic under standard conditions. The exceptions are the Noble Gases, which rarely form compounds. You simply have to know the so-called molecularity of oxygen, nitrogen, etc. Most inorganic chemists refer to these gases as #"dinitrogen"#, #"dioxygen"#, #"dihydrogen"# to avoid ambiguity.

Knowledge of the molecularity of the elemental gases would be assumed for an A level student, and it certainly would be expected of a 1st year college student.

Now chemical reactions follow experimental result; experimental result does not follow our depiction of the equation. For the reaction of oxygen and hydrogen, we know that 1/2 an equiv of dioxygen reacts with one equiv of dihydrogen under normal conditions to give one equiv of water:

#H_2(g) + 1/2O_2(g) rarr H_2O(l)# #(i)#

If I wanted to, I could double the entire equation to get rid of the half-integral coefficients.

#2H_2(g) + O_2(g) rarr 2H_2O(l)# #(ii)#

The point is that these equations are representations of chemical reality. Now these reactions certainly conserve mass: in #(i)# 18 g of dioxygen and dihydrogen combine to give 18 g of water (we would assume it to be a liquid under standard conditions!).

Given certain conditions, however, we could in fact produce hydrogen peroxide under so-called reducing conditions:

#H_2(g) + O_2(g) rarr HO-OH(l)# #(iii)#

This would be admittedly hard to do, nevertheless, we can still conceive of the reaction as a formality, and we can certainly estimate the thermodynamic parameters involved in the reaction. In the reaction #(iii)#, less energy would be evolved in the reaction because we are not completely reducing the oxygen gas, and we are only forming #2xxH-O# bonds rather than the #4xxH-O# bonds we form in #(ii)#. For another question dealing with the conservation of mass, see here.

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