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:

Options

A. `3.3xx10^(2) "mol" L^(-1)`
B. `3xx10^(-1) "mol" L^(-1)`
C. `3xx10^(-3) "mol" L^(-1)`
D. `3xx10^(3) "mol" L^(-1)`

Answer

C

Explanation

`C_([N_(2)O_(4))]=4.8xx10^(-2) molL^(-1), C_([NO_(2))]=1.2xx10^(-2) molL^(-1)`
`K_(C)=([NO_(2)]^(2))/([N_(2)O_(4)])=(1.2xx10^(-2)xx1.2xx10^(-3))/(4.8xx10^(-2))=0.3xx10^(-2)=3xx10^(-3) molL^(-1)`