For the reaction, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the concentration of an equilibrium mixture at `298 K` is `N_(2)O_(4)=4.50xx10^(-2) mol L^(-1)` and `NO_(2)=1.61xx10^(-2) mol L^(-1)`. What is the value of equilibrium constant?

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)`