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For an equilibrium reaction, N(2)O(4)(g)...

For an equilibrium reaction, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8 xx 10^(-2)` and `1.2 xx 10^(-2) mol//L` respectively. The value of `K_(c)` for the reaction is

A

`3.3xx10^(2)"mol litre"^(-1)`

B

`3xx10^(-1)"mol litre"^(-1)`

C

`3xx10^(-3)"mol litre"^(-1)`

D

`3xx10^(3)"mol litre"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`K=([NO_(2)]^(2))/([N_(2)O_(4)])=([1.2xx10^(-2)])/([4.8xx10^(-2)])=0.3xx10^(-2)=3xx10^(-3)`
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