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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


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

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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 $R=K[NO]^2$
answered Dec 18, 2013 by sreemathi.v
edited Jul 27, 2014 by balaji.thirumalai

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