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## order of reaction-

### Rate law ,

Rate = K [NO]p [Br2] q

initial rate ,

(Rate)0 = K [NO]0p [Br2] 0q       ———-eq 1

Rate law for experiment I,II,III-

Rate1 = K [0.10]p [0.10] q = 1.3 x 10 -6           ———eq 2

Rate2 = K [0.20]p [0.10] q  =5.2 x 10 -6           ———-eq3

Rate3 = K [0.20]p [0.30] q  = 1.56 x 10 -5         ——–eq4

comparing eq (2) and (3)

Rate 2 /rate 1 = K [0.20]p [0.10] q / K [0.10]p [0.10] q  =5.2 x 10 -6 / 1.3 x 10 -6

(2)p = 4

(2)p = (2)2

p = 2

comparing eq (3) and (4)

Rate 3 /rate 2 = K [0.20]p [0.30] q / K [0.20]p [0.10] q   = 1.56 x 10 -5 / 5.2 x 10 -6

(3)q = 3

(3)p = (3)1

q = 1

### Rate = K [A]p [B] q

Rate1 = K [0.10]p [1.0] q = 2.1 x 10 -3           ———eq 1

Rate2 = K [0.20]p [1.0] q  = 8.4 x 10 -3           ———-eq2

Rate3 = K [0.20]p [2.0] q  = 8.4 x 10 -3         ——–eq 3

comparing eq (2) and (1)

Rate 2 /rate 1 = K [0.20]p [1.0] q / K [0.10]p [1.0] q   = 8.4 x 10 -3 / 2.1 x 10 -3

(2)p = 4

(2)p = (2)2

p = 2

#### order of reaction with respect to ‘A’= 2

comparing eq (3) and (2)

Rate 3 /rate 2 = K [0.20]p [2.0] q / K [0.20]p [1.0] q   = 8.4 x 10 -3 / 8.4 x 10 -3

(2)q = 1

(2)q = (2)0

q = 0

### i) Rate law ,

Rate = K [Cl]p [NO] q

Rate law for experiment I,II,III-

Rate1 = K [0.05]p [0.05] q = 1.0 x 10 -3           ———eq 1

Rate2 = K [0.15]p [0.05] q  =3.0  x 10 -3           ———-eq2

Rate3 = K [0.05]p [0.15] q  = 9.0 x 10 -3            ——–eq 3

comparing eq (2) and (3)

Rate 2 /rate 1 = K [0.15]p [0.05] q / K [0.05]p [0.05] q   = 3.0 x 10 -3 / 1.0 x 10 -3

(3)p = 3

(3)p = (3)1

p = 1

comparing eq (3) and (1)

Rate 3 /rate 2 = K [0.05]p [0.15] q / K [0.05]p [0.05] q   = 9.0 x 10 -3 / 1.0 x 10 -3

(3)q = 9

(3)q = (3)2

q = 2

### iii) Rate = K [Cl2] [NO] 2

rate = 1 x 10-3

[NO] =0.05

[Cl2] =0.05

1 x 10 -3 = K x 0.05 x 0.05 x 0.05

K = 10-3 x 10 6 / 5 x 5 x 5

### iv) Rate = K [Cl2] [NO] 2

rate = ?

[NO] =0.4 M

[Cl2] =0.2 M

K =8.0 litre2 mole-2 sec-1

Rate  = 8 x 0.2 x 0.4 x 0.4

Rate = 0.256 mole litre-1 sec-1Ans.