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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 xx 10^(-3) mol//L`

B

`3.3 xx 10^(-3) mol//L`

C

`3 xx 10^(-1) mol//L`

D

`3.3 xx 10^(-1) mol//L`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)O_(4)(g) rarr 2NO_(2)(g) implies k_(c) = [NO_(2)]^(2)/[N_(2)O_(4)]`
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