The following data was obtained for chemical reaction given below at 975 K . \begin{tabular}{lclc} & $2 \mathrm{NO}_{(\mathrm{g})}+$ & $2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow$ & $\mathrm{N}_{2(\mathrm{g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$ \\ & {$[\mathrm{NO}]$} & {$\left[\mathrm{H}_{2}\right]$} & Rate \\ & $\mathrm{molL}^{-1}$ & $\mathrm{molL}^{-1}$ & $\mathrm{molL}^{-1} \mathrm{~s}^{-1}$ \\ (A) & $8 \times 10^{-5}$ & $8 \times 10^{-5}$ & $7 \times 10^{-9}$ \\ (B) & $24 \times 10^{-5}$ & $8 \times 10^{-5}$ & $2.1 \times 10^{-8}$ \\ (C) & $24 \times 10^{-5}$ & $32 \times 10^{-5}$ & $8.4 \times 10^{-8}$ \end{tabular} The order of the reaction with respect to NO is \_\_\_\_ . [Integer answer]

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Chemical Kinetics

The following data was obtained for chemical reaction given below at 975 K .

\begin{tabular}{lclc}
&

2 NO_{(g)} +

2 H_{2 (g)} \to

N_{2 (g)} + 2 H_{2} O_{(g)}

\\
& {

[NO]

} & {

[H_{2}]

} & Rate \\
&

molL^{- 1}

molL^{- 1}

molL^{- 1} s^{- 1}

\\
(A) &

8 \times 1 0^{- 5}

8 \times 1 0^{- 5}

7 \times 1 0^{- 9}

\\
(B) &

24 \times 1 0^{- 5}

8 \times 1 0^{- 5}

2.1 \times 1 0^{- 8}

\\
(C) &

24 \times 1 0^{- 5}

32 \times 1 0^{- 5}

8.4 \times 1 0^{- 8}

\end{tabular}

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

Answer: 1

Let the rate law be

rate = k [NO]^{x} [H_{2}]^{y}

. Between the first two sets of data the hydrogen concentration is unchanged while the nitric oxide concentration is tripled, so

\frac{2.1 \times 1 0 ^{- 8}}{7 \times 1 0 ^{- 9}} = (\frac{24 \times 1 0 ^{- 5}}{8 \times 1 0 ^{- 5}})^{x}

3 = 3^{x}

Hence

x = 1

, and the order of the reaction with respect to NO is

1

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer