`N_(2)O_(5)` decomposes according to equation,
`2N_(2)O_(5) rarr 4 NO_(2)+O_(2)`
(a) What does `-(d[N_(2)O_(5)])/(dt)` denote?
(b) What does `(d[O_(2)])/(dt)` denote?
(c) What is the unit of rate of this reaction?

Discuss the rate of the reaction 2N_(2)O_(5(g)) rarr 4NO_(2(g))+O_(2(g))

Discuss the rate of the reaction 2N_(2)O_(5(g)) rarr 4NO_(2(g))+O_(2(g))

Discuss the rate of the reaction 2N_(2)O_(5(g)) rarr 4NO_(2(g))+O_(2(g))

Discuss the rate of the reaction 2N_(2)O_(5(g)) rarr 4NO_(2(g))+O_(2(g))

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

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

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

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 ?