The decomposition of dinitrogen pentoxide is described by the chemical equation 2 N2O5(g) → 4 NO2(g) + O2(g) If the rate of disappearance of N2O5 is equal to 2.20 mol/min at a particular moment, what is the rate of appearance of NO2 at that moment?

Explanation:

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For the given reaction, the rate of disappearance of nitric oxide (negative as it is a reactant) has the following relationship with the rate of appearance of nitrogen dioxide, in terms of their stoichiometric coefficients:

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answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.