# Consider the reaction: 2 NO (g) + O2 (g) -->  2NO2  (g)  If at a particular moment during the reaction nitric oxide (NO) is reacting at a rate of 0.044 M/s, at what rate NO2 is being formed?  If at a particular moment during the reaction nitric oxide (NO) is reacting at a rate of 0.044 M/s, at what rate is O2 reacting?

In the given reaction:

`2NO (g) + O_2 (g) -> 2NO_2 (g)`

nitric oxide reacts with oxygen and forms nitrogen dioxide.

Using stoichiometry, 2 moles of nitric oxide reacts with 1 mole of oxygen and generates 2 moles of nitrogen dioxide.

The rate of nitric oxide reaction (or consumption) at a particular moment is given as 0.044 M/s. Since 2 moles of nitric oxide are forming 2 moles of nitrogen dioxide 0.044 M/s of nitrogen dioxide is being formed.

Since 2 moles of nitric oxide reacts with 1 mole of oxygen, the rate of reaction of oxygen would be half of nitric oxide. In other words, at that particular instant, oxygen would be reacting at 0.022 M/s (= 0.044/2 M/s).

Thus, using stoichiometry and a balanced chemical equation, we can determine the rate of reactions (consumption or formation) for various chemical compounds.

Hope this helps.

## See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Approved by eNotes Editorial Team