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The rate constant for the reaction, 2N(...

The rate constant for the reaction, `2N_(2)O_(5) to 4NO_(2) + O_(2)` is `2 xx 10^(-5) s^(-1)`. If rate of reaction is `1.4 xx 10^(-5) mol L^(-1), s^(-1)` What will be the concentration of `N_(2)O_(5)` in mol `L^(-1)`?

A

`1.4`

B

`1.2`

C

`0.04`

D

`0.8`

Text Solution

Verified by Experts

The correct Answer is:
D
Rate `= k[N_(2)O_(5)]`
`2.4 xx 10^(-5) mol L^(-1) s^(-1) = (3.0 xx 10^(-5) s^(-1)) [N_(2)O_(5)]`
`[N_(2)O_(5)] = (2.4 xx 10^(-5) mol L^(-1) s^(-1))/(3.0 xx 10^(-5) s^(-1)) = 0.8 mol L^(-1)`
Since the unit of `k` is `s^(-1)`, hence the decompoistion of `N_(2)O_(5)` is first order reaction.
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