CHEMISTRY TEXTBOOK

Problem 6.3 : For the reaction,

2 NOBr(g) 2 NO 2 (g) + Br 2 (g),

the rate law is rate = k[NOBr]^2. If the rate
of the reaction is 6.5 × 10-6 mol L-1 s-1 when
the concentration of NOBr is 2 × 10-3 mol
L-1. What would be the rate constant for the
reaction?

Solution :

rate = k[NOBr]^2 or k =

rate

[NOBr]^2

= 6.5 × 10

-6 mol L-1s-1

(2 × 10-3mol L-1)^2

= 1.625 mol-1 L-1 s-1

Try this...

• For the reaction

2A + 2B 2C +D, if

concentration of A is doubled at constant

[B] the rate increases by a factor of 4. If

the concentration of B is doubled with [A]

being constant the rate is doubled. Write

the rate law of the reaction.

• The rate law for the reaction

A + B C is found to be

rate = k[A]^2 [B].

The rate constant of the reaction at 25

(^0) C is 6.25 M-2s-1. What is the rate of
reaction when [A] = 1.0 mol dm-3 s-1 and
[B] = 0.2 mol dm-3 s -1?
6.3.3 Order of the reaction : For the reaction,
aA + bB cC + dD is
If the rate of the reaction is given as
rate = k[A]x[B]y.
Then the sum x + y gives overall
order of the reaction. Thus overall order of
the chemical reaction is given as the sum of
powers of the concentration terms in the rate
law expression. For example :
i. For the reaction,
2H 2 O 2 (g) 2H 2 O(l) + O 2 (g)
experimentally determined rate law is
rate = k[H 2 O 2 ].
The reaction is of first order.
ii. If the experimentally determined rate law
for the reaction
H 2 (g) + I 2 (g) 2 HI(g) is
rate = k[H 2 ][I 2 ].
The reaction is of first order in H 2 and I 2 each
and hence overall of second order.
Alternatively
The rate law gives rate = k [A]x[B]y.
a. From above observations (i) and (ii)
(i) 4 × 10-5 = (0.4)x(0.2)y
(ii) 6 × 10-5 = (0.6)x(0.2)y
Dividing (ii) by (i), we have
6 × 10-5
4 × 10-5 = 1.5 =
(0.6)x(0.2)y
(0.4)x(0.2)y =
0.6

(0.4

(x

= (1.5)x

Hence x = 1

b. From observations (i) and (iii) separately

in the rate law gives

iii) 4 × 10-5 = (0.4) × (0.2)y since x = 1

iv) 3.2 × 10-4 = 0.8 × (0.4)y

Dividing (iv) by (iii) we write

3.2 × 10

-4

4 × 10-5

=

0.8 (0.4) y

0.4 (0.2)y

or 8 = 2 × 0.4

(0.2

(y

or 4 = 2^2 = 2y

Therefore y = 2.

The rate law is then rate = k[A][B]^2.

Problem 6.2 contd....