Looking for classes? Ksquare Career Institute, Bengaluru →
The rate of the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways:
$\frac{-d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$
$\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between k and k′ and between k and k″ are
$\frac{-d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$
$\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between k and k′ and between k and k″ are
A
k′ = 2k; k″ = 2k
B
k′ = k; k″ = k
C
k′ = 2k; k″ = k
D
k′ = 2k; k″ = k/2
Detailed Solution
For $2N_2O_5 \rightarrow 4NO_2 + O_2$: Rate $= -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Substituting the three given expressions: $\frac{1}{2}k[N_2O_5] = \frac{1}{4}k'[N_2O_5] = k''[N_2O_5]$
$\frac{k}{2} = \frac{k'}{4} = k''$
From $\frac{k}{2} = \frac{k'}{4}$: $k' = 2k$
From $\frac{k}{2} = k''$: $k'' = \frac{k}{2}$
So k′ = 2k and k″ = k/2.
Substituting the three given expressions: $\frac{1}{2}k[N_2O_5] = \frac{1}{4}k'[N_2O_5] = k''[N_2O_5]$
$\frac{k}{2} = \frac{k'}{4} = k''$
From $\frac{k}{2} = \frac{k'}{4}$: $k' = 2k$
From $\frac{k}{2} = k''$: $k'' = \frac{k}{2}$
So k′ = 2k and k″ = k/2.