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

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

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

For a reaction, 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) rate of reaction is:

For a reaction, 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) rate of reaction is:

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 rate law of the reaction 2N_2O_5 rarr4NO_2+O_2 is

Gaseous N_(2)O_(5) decomposes according to the following equation: N_(2)O_(5)(g) rarr 2NO_(2)(g) + (1)/(2)O_(2)(g) The experimental rate law is -Delta [N_(2)O_(5)]//Delta t = k[N_(2)O_(5)] . At a certain temerature the rate constant is k = 5.0 xx 10^(-4) sec^(-1) . In how seconds will the concentration of N_(2)O_(5) decrease to one-tenth of its initial value ?

2N_(2)O_(5)(g)to4NO_(2)+O_(2)(g) What is the ratio of the rate of decomposition of N_(2)O_(5) to rate of formation of NO_(2)