\(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{s})\) This reaction was studied at \(-10^{\circ} \mathrm{C}\) and the following data was obtained \begin{tabular}{llll} run & {\([\mathrm{NO}]_{0}\)} & {\(\left[\mathrm{Cl}_{2}\right]_{0}\)} & \(\mathrm{r}_{0}\) \\ 1 & 0.10 & 0.10 & 0.18 \\ 2 & 0.10 & 0.20 & 0.35 \\ 3 & 0.20 & 0.20 & 1.40 \end{tabular} \([\mathrm{NO}]_{0}\) and \(\left[\mathrm{Cl}_{2}\right]_{0}\) are the initial concentrations and \(\mathrm{r}_{0}\) is the initial reaction rate. The overall order of the reaction is \_\_\_\_ . (Round off to the nearest integer).

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Chemical Kinetics

2 NO (g) + Cl_{2} (g) ⇌ 2 NOCl (s)

This reaction was studied at

- 1 0^{\circ} C

and the following data was obtained

\begin{tabular}{llll}
run & {

[NO]_{0}

} & {

[Cl_{2}]_{0}

} &

r_{0}

\\
1 & 0.10 & 0.10 & 0.18 \\
2 & 0.10 & 0.20 & 0.35 \\
3 & 0.20 & 0.20 & 1.40
\end{tabular}

[NO]_{0}

and

[Cl_{2}]_{0}

are the initial concentrations and

r_{0}

is the initial reaction rate. The overall order of the reaction is ____ . (Round off to the nearest integer).

Answer: 3

Write the rate law as

r_{0} = k [NO]^{m} [C l_{2}]^{n}

. Runs 1 and 2 hold

[NO]

fixed while

[C l_{2}]

is doubled, and the rate roughly doubles,

\frac{0.35}{0.18} \approx 2 = 2^{n}

, so

n = 1

. Runs 2 and 3 hold

[C l_{2}]

fixed while

[NO]

is doubled, and the rate becomes four times,

\frac{1.40}{0.35} = 4 = 2^{m}

, so

m = 2

. The overall order is

m + n = 3

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer