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The decomposition of N(2)O(5) takes plac...

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)`

Text Solution

Verified by Experts

(a) Rate `=K[N_(2)O_(5)]`
`:. K=(1.4xx10^(-6))/(0.04)=3.5xx10^(-5) sec^(-1)`
(b) Also `=K[N_(2)O_(5)]3.5xx10^(-5)xx1.20`
`4.2xx10^(-5) mol litre^(-1) sec^(-1)`
(c ) Further rate `=K[NO_(2)O_(5)]`
`:. [N_(2)O_(5)]=(2.45xx10^(-5))/(3.5xx10^(-5))`
`= 0.7 mol litre^(-1)`
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