#### Answer

Rate of disappearance of $NO_{2}$= 0.011 M/s.
Rate of the reaction= 0.0055 M/s.

#### Work Step by Step

Rate of the reaction=
$\frac{\Delta [N_{2}O_{5}]}{\Delta t}=-\frac{1}{2}\times\frac{\Delta [NO_{2}]}{\Delta t}= 0.0055 M/s$
$\implies \frac{\Delta [NO_{2}]}{\Delta t}= -2\times\frac{\Delta [N_{2}O_{5}]}{\Delta t}$
$=-2\times0.0055M/s= -0.011 M/s$.
Negative sign indicates the disappearance and the rate of disappearance of $NO_{2}$ is
0.011 M/s.