For the reaction,`2NO_(2)rightarrow2NO+O_(2)` ,rate is expressed as :
(a) `-(1)/(2)(d[NO_(2)])/(dt)=(1)/(2)(d[NO])/(dt)=(d[O_(2)])/(dt)`
(b) `-(1)/(2)(d[NO_(2)])/(dt)=-(1)/(2)(d[NO])/(dt)=(d[O_(2)])/(dt)`
(c) `-(2d[NO_(2)])/(dt)=(2d[NO])/(dt)=(d[O_(2)])/(dt)`
(d) `-(d[NO_(2)])/(dt)=(d[NO])/(dt)=(d[O_(2)])/(dt)`

The rate of reaction of certain reaction is expressed as : (1)/(3)(d[A])/(dt)=-(1)/(2)(d[B])/(dt)=-(d[C])/(dt) The reaction is :

For the reaction, 2NOrarrN_(2)+O_(2) , the expression -(1)/(2)(d[NO])/(dt) represents:

(d)/(dt)log(1+t^(2))

Consider the decomposition of N_(2)O_(5) as N_(2)O_(5)rarr2NO_(2)+1//2O_(2) The rate of reaction is given by -(d[N_(2)O_(5)])/(dt)=(1)/(2)(d[NO_(2)])/(dt)=2(d[O_(2)])/(dt) =k_(1)[N_(2)O_(5)] Therefore (-d[N_(2)O_(5)])/(dt)=k_(1)[N_(2)O_(5)] (+d[NO_(2)])/(dt)=2k_(1)[N_(2)O_(5)]=k_(1)[N_(2)O_(5)] (+d[O_(2)])/(dt)=(1)/(2)k_(1)[N_(2)O_(5)]=k_(1)[N_(2)O_(5)] Choose the correct option

For the reaction 2N_(2)O_(4)iff 4NO_(2) , given that (-d[N_(2)O_(4)])/(dt)=K " and "(d[NO_(2)])/(dt)=K , then

The rate of a reaction is expressed in different ways as follows: +(1)/(2)(d[C])/(dt)=-(1)/(3)(d[D])/(dt)=+(1)/(4)(d[A])/(dt)=-(d[B])/(dt) the reaction is