Use the data to determine rate law for reaction: $2NO+H_2\rightarrow N_2O+H_2O$

 Experiment No $[NO]$ $[H_2]$ Initial rate 1 0.021 0.065 1.46M/min 2 0.021 0.260 1.46M/min 3 0.042 0.065 5.84M/min

$\begin{array}{1 1}(a)\;R=K[NO][H_2]\\(B)\;R=K[NO]^2[H_2]\\(c)\;R=K[H_2]^2\\(d)\;R=K[NO]^2\end{array}$

$R=K[NO]^x[H_2]^y$
If $[NO]$ is kept constant if we vary $[H_2]$ then rate is constant
So $y=0$
If $[H_2]$ kept constant and [NO] doubled then rate becomes 4 times .
So $x=2$
So $R=K[NO]^2$
edited Jul 27, 2014