`2N_(2)O_(5)to4NO_(2)+O_(2)`. The rate of reaction in terms of `N_(2)O_(5)` will be

For the reaction : 2N_(2)O_(5) rarr 4NO_(2) +O_(2) , the rate of reaction in terms of O_(2) is d[O_(2)] /dt. In terms of N_(2)O_(5) , it will be :

For the reaction : 2N_(2)O_(5) rarr 4NO_(2) +O_(2) , the rate of reaction in terms of O_(2) is d[O_(2)] /dt. In terms of N_(2)O_(5) , it will be :

The decomposition of nitrogen pentoxide is given as 2N_(2)O_(5) to 4NO_(2) + O_(2) . The rates of reaction are ([N_(2)O_(5)])/(Delta t ) =k_(1)[N_(2)O_(5)],(Delta[NO_(2)])/(Deltat) = k_(2) [N_(2)O_(5)] and (Delta [O_(2)])/(Delta t) = k_(3) [N_(2)O_(5)] Relate the rate constants k_(1) , k_(2) and k_(3) .

For the reaction 2N_(2)O_(5)rarr4NO_(2)+O_(2) the rate equation can be express as (d[N_(2)O_(5)])/(dt)=k[N_(2)O_(5) and (d[NO_(2)])/(dt)=k'[N_(2)O_(5)] k and k' are related as

The decomposition of N_(2)O_(5) takes place according to I order as: 2N_(2)O_(5) rarr 4NO_(2)+O_(2) Calculate: (a) The rate constant, if instantaneous rate is 1.4 xx 10^(-6) mol litre^(-1) sec^(-1) when concentration of N_(2)O_(5) is 0.04M . (b) The rate of reaction when concentration of N_(2)O_(5) is 1.20 M . (c ) The concentration of N_(2)O_(5) when the rate of reaction will be 2.45 xx 10^(-5) mol litre^(-1) sec^(-1)

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 a reaction , 2N_(2)O_(5)rarr4NO_(2)+O_(2), the rate is directly proportional to [N_(2)O_(5)]. At 45^(@)C, 90% of the N_(2)O_(5) react in 3600 s. The value of the rate constant is