Question

Given that 2NO2→2NO+O2
If the rate of decrease in concentration of NO2 is 4.0×10-¹³ S-¹, calculate the rates of increase in concentration of NO and NO2.​

Answer 1

Answer:

Here rate law is given as:-

rate=k[NO]

2

[O

2

]........(1)

[NO](concentration of NO) can be written as

V

n

where n=number of moles and V=volume

From (1)

rate=K[

V

n

]

2

[

V

n

]......(2)

Now,

Volume is related to 1/3

r

d

From V to

3

V

put new volume in (2)

New rate will be

r

′

=K[

V

3n

]

2

[

V

3n

]=K(3)

2

[

V

n

]

2

(2)[

V

n

]=27K[

V

3n

]

2

[

V

3n

]

so,

r

r

′

=

K[

V

n

]

2

[

V

n

]

27K[

V

3n

]

2

[

V

3n

]

=27

Rate of reaction changes by 27 times. plz follow me and like this answer