Question

Consider the reaction 2 NO + O2 → 2 NO2

Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.050 M/s.
(a) At what rate is NO2 being formed?
(b) At what rate is molecular oxygen reacting?

Answer 1

(a) Rate at which
`NO_2` is formed is 0.050 M/s

(b) Rate at which
`O_2` is consumed is 0.0250 M/s.

Step-by-step explanation:

The given reaction is:-

`2NO+O_2\rightarrow 2NO_2`

The expression for rate can be written as:-

`-(1)/(2)(d[NO])/(dt)=-(d[O_2])/(dt)=(1)/(2)(d[NO_2])/(dt)`

Given that:-
`(d[NO])/(dt)=-0.050\ M/s` (Negative sign shows consumption)

`-(1)/(2)(d[NO])/(dt)=(1)/(2)(d[NO_2])/(dt)`

`-(d[NO])/(dt)=(d[NO_2])/(dt)`

`-(-0.050\ M/s)=(d[NO_2])/(dt)`

`(d[NO_2])/(dt)=0.050\ M/s`

(a) Rate at which
`NO_2` is formed is 0.050 M/s

`-(1)/(2)(d[NO])/(dt)=-(d[O_2])/(dt)`

`-(1)/(2)* -0.050\ M/s=-(d[O_2])/(dt)`

`(d[O_2])/(dt)=0.0250\ M/s`

(b) Rate at which
`O_2` is consumed is 0.0250 M/s.