Question

\[\ce{2NO_{(g)} + Cl2_{(g)} <=> 2NOCl_{(s)}}\]

This reaction was studied at −10°C and the following data was obtained

Run	[NO]0	[Cl2]0	r0
1	0.10	0.10	0.18
2	0.10	0.20	0.35
3	0.20	0.20	1.40

[NO]₀ and [Cl₂]₀ are the initial concentrations and r₀ is the initial reaction rate. The overall order of the reaction is ______ (Round off to the nearest integer).

Answer 1

[NO]₀ and [Cl₂]₀ are the initial concentrations and r₀ is the initial reaction rate. The overall order of the reaction is 3.

Explanation:

The given reaction is

\[\ce{2NO_{(g)} + Cl2_{(g)} <=> 2NOCl_{(s)}}\]

Rate law r = k[NO]^x[Cl2]^y

Compare Run 1 and Run 2:

[NO] is constant = 0.10

[Cl₂] doubles (0.10 → 0.20)

Rate changes: 0.18 → 0.35

\[\ce{\frac{r_2}{r_1} = \frac{0.35}{0.18}}\]

= 1.94

= 2

So, when [Cl₂] doubles, rate doubles → order in Cl₂ (y) = 1

Compare Run 2 and Run 3

[Cl₂] is constant = 0.20

[NO] doubles (0.10 → 0.20)

Rate changes: 0.35 → 1.40

`r_3/r_2 = 1.40/0.35`

= 4

So, when [NO] doubles, rate increases 4 times → order in NO (x) = 2

Overall order = x + y = 2 + 1 = 3