# The following data was obtained for chemical reaction given below at $975 mathrm{~K}$.

Question:

The following data was obtained for chemical reaction given below at $975 \mathrm{~K}$.

$2 \mathrm{NO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$

$\begin{array}{lcl}{[\mathrm{NO}]} & {\left[\mathrm{H}_{2}\right]} & \text { Rate } \\ \text { mol } \mathrm{L}^{-1} & \mathrm{~mol} \mathrm{~L}^{-1} & \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\end{array}$

(A) $8 \times 10^{-5} \quad 8 \times 10^{-5} \quad 7 \times 10^{-9}$

(B) $24 \times 10^{-5} \quad 8 \times 10^{-5} \quad 2.1 \times 10^{-8}$

(C) $24 \times 10^{-5} \quad 32 \times 10^{-5} \quad 8.4 \times 10^{-8}$

The order of the reaction with respect to NO is - [Integer answer]

Solution:

$7 \times 10^{-9}=\mathrm{K} \times\left(8 \times 10^{-5}\right)^{\mathrm{x}}\left(8 \times 10^{-5}\right)^{\mathrm{y}} \quad \ldots \ldots .(1)$

$2.1 \times 10^{-8}=\mathrm{K} \times\left(24 \times 10^{-5}\right)^{\mathrm{x}}\left(8 \times 10^{-5}\right)^{\mathrm{y}} \quad \ldots \ldots .(2)$

$\frac{1}{3}=\left(\frac{1}{3}\right)^{x} \Rightarrow x=1$