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The rate constant for the reaction : 2...

The rate constant for the reaction :
`2N_(2) O_(5) rarr 4NO_(2) +O_(2)`
is `3.0xx10^(-5) "sec"^(-1)` . If the rate is `2.40xx10^(-5)` mol `"litre"^(-1) "sec"^(-1)` then the concentration of `N_(2)O_(5)` (in mol `"litre"^(-1)` ) is :

A

`1.4`

B

`1.2`

C

`0.02`

D

`0.8`

Text Solution

Verified by Experts

The correct Answer is:
D
(D) Rate`= k[N_(2)O_(5)]`
or `[N_(2)O_(5)] = ("Rate")/(k)`
`= (2.40xx10^(-5))/(3.0xx10^(-5))` =0.8
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