Answer
a) $Rate= k[NO]^{2}[Cl_{2}]$
b) $180\, L^{2}/mol^{2}\cdot min$
Work Step by Step
a) $Rate= k[NO]^{x}[Cl_{2}]^{y}$
When the concentration of NO is doubled and that of $Cl_{2}$ is kept constant, then the initial rate increases by a factor of four from 0.36 to 1.45
This implies that the rate depends upon the square of [NO].
When the concentration of $Cl_{2}$ is doubled and that of NO is kept constant, the rate also gets doubled.
This implies that the rate depends on the first power of $[Cl_{2}]$.
Hence, $Rate= k[NO]^{2}[Cl_{2}]$
b) $k= \frac{Rate}{[NO]^{2}[Cl_{2}]}=\frac{0.18\,mol/L\cdot min}{(0.10\,mol/L)^{2}\times0.10\,mol/L}=180\, L^{2}/mol^{2}\cdot min$
