Respuesta :
Answer:
(a) Rate at which [tex]NO_2[/tex] is formed is 0.050 M/s
(b) Rate at which [tex]O_2[/tex] is consumed is 0.0250 M/s.
Explanation:
The given reaction is:-
[tex]2NO+O_2\rightarrow 2NO_2[/tex]
The expression for rate can be written as:-
[tex]-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}[/tex]
Given that:- [tex]\frac{d[NO]}{dt}=-0.050\ M/s[/tex] (Negative sign shows consumption)
[tex]-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}[/tex]
[tex]-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}[/tex]
[tex]-(-0.050\ M/s)=\frac{d[NO_2]}{dt}[/tex]
[tex]\frac{d[NO_2]}{dt}=0.050\ M/s[/tex]
(a) Rate at which [tex]NO_2[/tex] is formed is 0.050 M/s
[tex]-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}[/tex]
[tex]-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}[/tex]
[tex]\frac{d[O_2]}{dt}=0.0250\ M/s[/tex]
(b) Rate at which [tex]O_2[/tex] is consumed is 0.0250 M/s.
