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For the reaction 2N(2)O(5) rarr 4NO(2)+O...

For the reaction `2N_(2)O_(5) rarr 4NO_(2)+O_(2)` rate of reaction and rate constant are `1.02 xx 10^(-4)` and `3.4 xx 10^(-5) sec^(-1)` respectively. The concentration of `N_(2)O_(5)` at that time will be

A

`1.732 "mol L"^(-1)`

B

`3 "mol L"^(-1)`

C

`1.02 xx 10^(-4) "mol L"^(-1)`

D

`3.2 xx 10^(5) "mol L"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
From units of rate constant `(s^(-1))`, the order of reaction = 1
`therefore` rate `= k [N_(2)O_(5)]`
`1.02xx10^(-4)=3.4xx10^(-5)[N_(2)O_(5)]`
`[N_(2)O_(5)]=(1.02xx10^(-4))/(3.4)xx10^(-5)=3 mol L^(-1)`
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