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The rate constant for the reaction: 2N...

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) M sec^(-1)`, then the concentration of `N_(2)O_(5)` (in M) is:`

A

1.4

B

1.2

C

0.04

D

0.8

Text Solution

Verified by Experts

The correct Answer is:
D
`k = 3.0 xx 10^(-5) s^(-1)`
rate `= 2.4 xx 10^(-5) "mol L"^(-2)s^(-1)`
As units of k is `s^(-1)`, therefore, the reaction is of first order
`:.` rate = `k[N_(2)O_(5)]`
`2.4 xx 10^(-5) "mol"^(-1) L^(-1)s^(-1)=(3.0 xx 10^(-5)s^(-1))[N_(2)O_(5)]`
`[N_(2)O_(5)](2.4xx10^(-5)"mol L"^(-1)s^(-1))/(3.0xx10^(-5)s^(-1))=0.8"mol L"^(-1)`.
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