Question

2 NO(g) + O2(g) → 2 NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.070 M/s. (a) At what rate is NO2 being formed? (b) What rate is molecular oxygen reacting?

Answer 1

Consider,

`2NO(g) + O_2(g) to 2NO_2(g)`

Moreover, understand that reaction rates are proportional to the stoichiometry of the reaction. Hence,

`(0.070M)/s * (2NO_2)/(2NO) = (0.070M)/s`

is the rate of appearance of nitrogen dioxide, and

`(0.070M/s) * (O_2)/(2NO) = (-0.035M)/s`

is the rate of disappearance of the oxygen.

To be sure, reaction rates are merely,

`(Delta[A])/(Deltat)`

and the concentrations of the constituents are proportional to the stoichiometry.

If I made a mistake, I'm open to feedback!