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The decomposition of `N_(2) O_(5)` is represented by the equation `2N_(2)O_(5(g)) to 4NO_(2(g)) + O_(2(g))` (a ) How is the rate of formation of `NO_(2)` related to the rate of formation of `O_(2)`? (b ) How is the rate of formation of `O_(2)` related to the rate of consumption of `N_(2)O_(s)?` |
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Answer» Given : `2N_(2)O_(5(g)) to 4NO_(2(g)) + O_(2(g))` (a) Rate of formation of `O_(2)` at time `t= (d[O_(2)])/(dt)` They are related to each other through rate of reaction `:. ` Rate of reaction `=(1)/(4) (d[NO_(2)])/(dt)= (d[O_(2)])/(dt)` ( b) Rate of consumption of `N_(2) O_(5)` at time `t= - (d[N_(2)O_(5)])/(dt)` Rate of reaction `=- (1)/(2) (d[N_(2)O_(5)])/(dt) =(d[O_(2)])/(dt)` In general Rate of reaction `= -(1)/(2) (d[N_(2)O_(5)])/(dt)= (1)/(4) (d[NO_(2)])/(dt)= (d[O_(2)])/(dt)` |
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