In the equation #2HI(l) -> H_2(g) + I_2(s)#, what are the reactants?

1 Answer
anor277
Mar 16, 2016

As written, hydroiodic acid is the reactant.

Explanation:

Of course, this is an equilibrium reaction:

#2HI(g) rightleftharpoons I_2(g) + H_2(g)#, in which there is equality of forward and reverse rates.

I could equally write,

#I_2(g) + H_2(g) rightleftharpoons 2HI(g)#,

in which we would regard the elements as the reactants.

Which side you designate as reactants depends on whether you are interested the DECOMPOSITION of hydrogen iodide, or the FORMATION of hydrogen iodide from its elements.

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