| The rate of reaction \[2{{N}_{2}}{{O}_{5}}\xrightarrow[{}]{{}}4N{{O}_{2}}+{{O}_{2}}\]can be written in three ways |
| \[\frac{-d[{{N}_{2}}{{O}_{5}}]}{dt}=k[{{N}_{2}}{{O}_{5}}]\] \[\frac{d[NO_{7}^{2}]}{dt}=k'[{{N}_{2}}{{O}_{5}}]\] |
| \[\frac{d[O_{2}^{{}}]}{dt}=k''[{{N}_{2}}{{O}_{5}}]\] [AIPMT (M) 2011] |
| The relationship between k and k' and between k and k'' are |
A) k = 2 k; k = k
B) k = 2k; k=k/2
C) k = 2k; k = 2k
D) k = k; k = k
Correct Answer: B
Solution :
| [b] Rate \[=-\frac{1}{2}\frac{d[{{N}_{2}}{{O}_{5}}]}{dt}=\frac{1}{4}\frac{d[N{{O}_{2}}]}{dt}=\frac{d[{{O}_{2}}]}{dt}\] |
| \[\Rightarrow \]\[\frac{1}{2}k[{{N}_{2}}{{O}_{5}}]=\frac{1}{4}k'[{{N}_{2}}{{O}_{5}}]=k'\,'[{{N}_{2}}{{O}_{5}}]\] |
| \[\Rightarrow \] \[\frac{k}{2}=\frac{k'}{4}=k'\,'\] |
| \[\therefore \] \[k'=2k;k'\,'=\frac{k}{2}\] |
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