The decomposition of N(2)O(5) occurs as,...

The decomposition of `N_(2)O_(5)` occurs as, `2N_(2)O_(5) rarr4NO_(2)+O_(2)` and follows `I` order kinetics, hence:

The reaction is bimolecular

The reaction is unimolecular

`t_(a) propa^(@)`

None of these

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The correct Answer is:

For `I^(st)` order reaction half life is independent of concentration.

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The thermal decomposition of N_(2)O_(5) occues as: 2N_(2)O_(5) rarr 4NO_(2)+O_(2) Experimental studies suggest that rate of decomposition of N_(2)O_(5) rate of formation of NO_(2) or rate of formation of O_(2) all becomes double if concentration of N_(2)O_(5) is doubled. If rate of formation of O_(2) is 16 g//hr then rate of decomposition of N_(2)O_(5) and rate of formation of NO_(2) respectively is

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

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