Question

The rate of the reaction 2N₂O₅ → 4NO₂ + 2O₂ can be written
in three ways :
- d[N₂O₅ ]/dt = k [N₂O₅ ]
d[NO₂ ]/dt = k' [N₂O₅ ]
d[O₂ ]/dt = k'' [N₂O₅ ]
The relationship between k and k' and between k and k¢¢ are:
(a) k' = 2k ; k' = k (b) k' = 2k ; k'' = k / 2
(c) k' = 2k ; k'' = 2k (d) k' = k ; k'' = k

Answer 1

answer : option (a) k' = 2k ; k" = k

The rate of the reaction 2N₂O₅ → 4NO₂ + 2O₂ can be written in three ways :

1. - d[N₂O₅ ]/dt = k [N₂O₅ ]
2. d[NO₂ ]/dt = k' [N₂O₅ ]
3. d[O₂ ]/dt = k'' [N₂O₅ ]

but from reaction,

-1/2 d[N₂O₅]/dt = 1/4 d[NO₂]/dt = 1/2 d[O₂]/dt .......(i)

now from equation (1) and (2),

{-d[N₂O₅]/dt }/{d[NO₂]/dt} = k[N₂O₅]/k'[N₂O₅]

from equation (i)

⇒1/2 = k/k'

⇒k' = 2k ........(ii)

again from equation (1) and (3),

{-d[N₂O₅]/dt}/{d[O₂]/dt} = k[N₂O₅]/k"[N₂O₅]

⇒1 = k/k"

⇒k" = k........(iii)

we see, only option (a) follow the conditions (ii) and (iii). hence, option (a) is correct choice.

Answer 2

Answer:

answer: option (a) k' = 2k;

k" = k

The rate of the reaction 2N₂O; + 4NO₂ +20, can be written in three ways:

1.- d[N₂O₁]/dt = k [N₂O, ]

2. d[NO₂ ]/dt-k' [N₂O₁]

3. d[O₂ /dt k" [N₂O₂]

but from reaction,

1/2 d[N₂O.]/dt = 1/4 d[NO₂]/dt = 1/2 d[0₂]

now from equation (1) and (2), [-d[N₂O,]/dt [d[NO₂]/dt] = K[N₂O₂]/

from equation (1)

-1/2 = k/k

ak' 2k.

again from equation (1) and (3).

-1=k/k"

k" = (ii)

we see, only option (a) follow the conditions (ii) and (iii), hence, option (a) is correct choice.