2NO(g) + Cl2(g) → 2NOCl(g) The following data were collected. All the measurements were taken at 263 K. Experiment No. Initial [NO] (M) Initial [Cl2] (M) Initial rate of disappearance of Cl2 (M/min) 1 0.15 0.15 0.60 2 0.15 0.30 1.20 3 0.30 0.15 2.40 4 0.25 0.25 ?

(a) Rate law may be written as
Rate = k[NO]p[Cl2]p
The initial rate becomes
(Rate)0 = k[NO]p[Cl2]p
Comparing experiment 1 and 2
(Rate)1 = k(0.15)p(0.15)p = 0.60.. (i)
(Rate)2 = k(0.15)p(0.30)p = 1.20.. (ii)
Dividing equation (ii) by (i)
(Rate)2/(Rate)2 = k(0.15)p(0.30)p/k(0.15)p(0.1)p = 1.20/0.60
or 2q = 2
1q = 1
order with respect to Cl2 = 1
Comparing experiment 1 and 3
(Rate)1 = k(0.15)p(0.15)q = 0.60.. (i)
(Rate)3 = k(0.30)p(0.15)q = 2.40.. (ii)
Dividing equation (ii) by (i)
(Rate)3/(Rate)1 = k(0.30)p(0.15)q/k(0.15)p(0.15)q = 2.40/0.60
or 2r = 4
2q = 22
2p = 2
Thus order with respect to NO is 2
Rate law = k[NO]2[Cl2]1
(b) The rate law for the reaction Rate = k[NO]2[Cl2]
Rate constant can be calculated by substituting the value of rate, [NO] and [Cl2] for any experiment
k = Rate/[NO]2[Cl2] = 0.60/(0.15)2(0.15) = 0.60/0.003375 = 177.77 mol⁻²L²min⁻¹
(c) Let initial rate of disappearance of Cl2 in exp. 4 is r4
∴r4 = k[NO]2[Cl2] = 177.77 × (0.25)2(0.25)