Answer
(a) 0.066 M/s
(b) -0.033 M/s
Work Step by Step
From the reaction, rate of the reaction, $ R$ is
$$ R = \frac{-d[NO]}{2dT} = \frac{-d[O_2]}{dT} = \frac{d[NO_2]}{2dT}$$
It is given that
$$ \frac{-d[NO]}{dT} = 0.066 M/s$$
Therefore, $$ R= \frac{0.066}{2} = 0.033 M/s$$
(a) Rate of formation of $NO_2$ is given by
$$\frac{d[NO_2]}{dT} = 2\times R = 2\times0.033 M/s = 0.066 M/s$$
(b) Rate of reaction of molecular oxygen is given by,
$$ \frac{d[O_2]}{dT} = -R = -0.033 M/s $$
