Answer
See the answer below.
Work Step by Step
a) From the table, doubling the concentration of NO, at the same O2 concentration, leads to the reaction rate being quadrupled, so the reaction is of order 2 in NO. Doubling the concentration of O2, at the same NO concentration, leads to the reaction rate being doubled, so the reaction is of order 1 in O2.
b) $r=k[NO]^2[O_2]$
c) From the first line of the table:
$2.5\times10^{-5}\ mol/L.s=k(0.01\ mol/L)^2(0.01\ mol/L)$
$k=25.0\ L^2/mol^2.s$
d) $r=25\times(0.015)^2(0.005)=2.81\times10^{-5}\ mol/L.s$
e) From stoichiometry, NO2 is being formed at $1.0\times10^{-4}\ mol/L.s$ and O2 is reacting at $0.5\times10^{-4}\ mol/L.s$
