order of reaction
source : youtube
order of reaction-
Q 1) 2 NO + Br2 ———> 2 NOBr
Experiment initial conc. initial rate ( mole litre-1 min -1)
[NO] [Br2]
I 0.10 0.10 1.3 x 10 -6
II 0.20 0.10 5.2 x 10 -6
III 0.20 0.30 1.56 x 10 -5
Determine
i) the orders with respect to NO and Br2.
ii) the rate law
iii) rate constant
Solution )
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
order of reaction = p + q = 2 + 1 = 3 Ans.
Q 2) Consider the following data for the reaction-
A + B ——-> Products
Experiment initial conc.’A’ initial conc.’B’ initial rate
I 0.10 M 1.0M 2.1 x 10 -3
II 0.20 M 1.0 M 8.4 x 10-3
III 0.20 M 2.0 M 8.4 x 10 -3
Determine the orders with respect to A and with respect to B and overall order of reaction?
Solution )
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
order of reaction with respect to ‘B’= 0
order of reaction = p + q = 2 + 0 = 2 Ans.
Q 3) 2 NO + Cl2 ———> 2 NOCl at 295 K
Experiment initial conc. initial rate ( mole litre-1 min -1)
[NO] [Cl2]
I 0.05 0.05 1.0 x 10 -3
II 0.15 0.05 3.0 x 10 -3
III 0.05 0.15 9.0 x 10 -3
Determine
i) What is the order with respect to NO and Cl2 in the reaction?
ii) Write the rate law or expression.
iii) Calculate rate constant.
iv) Calculate reaction rate when concentrations of Cl2 and NO are 0.2 M and 0.4 M respectively ?
Solution )
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
Order of reaction with respect to Cl2 = 1
Order of reaction with respect to NO = 2
Overall order of reaction = p + q = 1 + 2 = 3 Ans.
ii) Rate = K [Cl2] [NO] 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
K =8.0 litre2 mole-2 sec-1
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.