Question

Consider the reaction 2 NO + O2 → 2 NO2

Answer 1

Answer:

(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.

Explanation:

The given reaction is:-

$2NO+O_2\rightarrow 2NO_2$

The expression for rate can be written as:-

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

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

$-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

$-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}$

$-(-0.050\ M/s)=\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=0.050\ M/s$

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

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}$

$-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}$

$\frac{d[O_2]}{dt}=0.0250\ M/s$

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