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

For the reaction equilibrium, `N_(2)O_(4(g))hArr2NO_(2(g))`, the concentration of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8xx10^(-2)` and `1.2xx10^(-2)` mol/L respectively. The value of `K_(c)` for the reaction is:

A

`3xx10^(-3)`M

B

`3xx10^(3)`M

C

`3.3xx10^(2)`M

D

`3xx10^(-1)`M

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(c) = ([NO_(2)]^(2))/([N_(2)O_(4)]) = ((1.2xx10^(-2))^(2))/(4.8xx10^(-2)) = 3.0xx10^(-3)` M
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