SATHEE: Chemical Kinetics - Result Question 7

Chemical Kinetics - Result Question 7

####7. The rate of the reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ can be written in three ways :

[2011 M]

$\frac{- d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]$

$\frac{d [N O_{2}]}{d t} = k^{'} [N_{2} O_{5}]$

$\frac{d [O_{2}]}{d t} = k^{''} [N_{2} O_{5}]$

The relationship between $k$ and $k^{'}$ and between $k$ and $k^{''}$ are :

(a) $k^{'} = 2 k; k^{'} = k$

(b) $k^{'} = 2 k; k^{''} = k / 2$

(d) $k^{'} = k; k^{''} = k$

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Solution:

(b) Rate of disappearance of reactants = Rate of appearance of products

$- \frac{1}{2} \frac{d (N_{2} O_{5})}{d t} = \frac{1}{4} \frac{d (N O_{2})}{d t} = \frac{d (O_{2})}{d t}$

$\frac{1}{2} k (N_{2} O_{5}) = \frac{1}{4} k^{'} (N_{2} O_{5}) = k^{''} (N_{2} O_{5})$

$\frac{k}{2} = \frac{k^{'}}{4} = k^{''}$

$k^{'} = 2 k, k^{''} = \frac{k}{2}$

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Chemical Kinetics - Result Question 7

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