2NO(g) + Cl 2 (g) leftharpoons 2NOCl(s) This reaction was studied at - 10 ^\circ and the following data was obtained 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 [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 \underline\hspace2cm . (Round off to the Nearest Integer).

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2021 · 18 Mar · Shift 1 · Q17 with Solution

2NO(g) + Cl₂(g)

\rightleftharpoons

2NOCl(s)

This reaction was studied at

-

^\circ

and the following data was obtained

Run	${[NO]_0}$	${[C{l_2}]_0}$	${r_0}$
1	0.10	0.10	0.18
2	0.10	0.20	0.35
3	0.20	0.20	1.40

{[NO]_0}

and

{[C{l_2}]_0}

are the initial concentrations and r₀ is the initial reaction rate.

The overall order of the reaction is

\underline{\hspace{2cm}}

. (Round off to the Nearest Integer).

Numerical answer

View written solutionFree

Correct answer: 3

Assume the rate law

Let the initial rate be $r = k[NO]^m[Cl_2]^n$ where $m$ and $n$ are the orders with respect to $NO$ and $Cl_2$ .
Use Runs 1 and 2 to find order in $Cl_2$

From the data:
- Run 1: $[NO]=0.10,\, [Cl_2]=0.10,\, r=0.18$
- Run 2: $[NO]=0.10,\, [Cl_2]=0.20,\, r=0.35$
Here, $[NO]$ is constant and $[Cl_2]$ is doubled.

So, $\frac{r_2}{r_1} = \left(\frac{[Cl_2]_2}{[Cl_2]_1}\right)^n$ $\frac{0.35}{0.18} = \left(\frac{0.20}{0.10}\right)^n$ $1.944 \approx 2^n$

Since $1.944 \approx 2$ , we get $n \approx 1$
Use Runs 2 and 3 to find order in $NO$

From the data:
- Run 2: $[NO]=0.10,\, [Cl_2]=0.20,\, r=0.35$
- Run 3: $[NO]=0.20,\, [Cl_2]=0.20,\, r=1.40$
Here, $[Cl_2]$ is constant and $[NO]$ is doubled.

So, $\frac{r_3}{r_2} = \left(\frac{[NO]_3}{[NO]_2}\right)^m$ $\frac{1.40}{0.35} = \left(\frac{0.20}{0.10}\right)^m$ $4 = 2^m$

Hence, $m = 2$
Find overall order

Overall order $= m+n = 2+1 = 3$
Rounded to nearest integer

$\boxed{3}$