Question

For the reaction

\[\ce{H2O2_{(aq)} + 2H{^+_{(aq)}} + 2I{^-_{(aq)}} -> I2_{(aq)} + H2O}\],

following data are obtained:

Expt.	[H2O2]	[I−]	[H+]	d[I2]/dt
I	0.01	0.01	0.10	1.75 × 10−6
II	0.03	0.01	0.10	5.25 × 10−6
III	0.03	0.02	0.10	1.05 × 10−5
IV	0.03	0.02	0.20	1.05 × 10−5

Determine the rate law and calculate the value of rate constant.

Answer 1

Suppose the orders of the given reaction with respect to H₂O₂, I⁻ and H⁺ are p, q and r, respectively. The rate law can thus be written as

(For set I): 1.75 × 10⁻⁶ = k (0.01)^p (0.01)^q (0.10)^r    ...(i)

(For set II): 5.25 × 10⁻⁶ = k (0.03)^p (0.01)^q (0.10)^r    ...(ii)

(For set III): 1.05 × 10⁻⁵ = k (0.03)^p (0.02)^q (0.10)^r    ...(iii)

(For set IV): 1.05 × 10⁻⁵ = k (0.03)^p (0.02)^q (0.20)^r    ...(iv)

Dividing eq. (i) by eq. (ii), we get

`(1.75 xx 10^-6)/(5.25 xx 10^-6) = (0.01/0.03)^p`

or, `(1/3)^1 = (1/3)^p`

∴ p = 1

Dividing eq. (ii) by eq. (iii), we get

`(5.25 xx 10^-6)/(1.05 xx 10^-5) = (0.01/0.02)^q`

or, `(1/2)^1 = (1/2)^q`

∴ q = 1

Dividing eq. (iii) by eq. (iv), we get

`(1.05 xx 10^-5)/(1.05 xx 10^-5) = (0.10/0.20)^r`

or, 1 = `(1/2)^r`

∴ r = 0

Hence, the rate law is

Rate = k [H₂O₂]¹ [I⁻]¹ [H⁺]⁰

or, Rate = k [H₂O₂] [I⁻]

From eq. (i) we have,

k = `(1.75 xx 10^-6)/((0.01)^1 xx (0.01)^1 xx (0.10)^0)`

= 1.75 × 10⁻² mol⁻¹ L s⁻¹.