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

Options

A. `3.3 xx 10^(2) mol L^(-1)`
B. `3.3xx10^(-1) mol L^(-1)`
C. `3.3 xx 10^(-3) mol L^(-1)`
D. `3.3 xx 10^(3) mol L^(-1)`

Answer

C

Explanation

`N_(2)O_(4)(g) hArr 2NO_(2)(g)`
`K_(c)=([NO_(2)]^(2))/([N_(2)O_(4)])=((1.2xx10^(-2)"mol"L^(-1))^(2))/(4.8xx10^(-2)"mol"L^(-1))`
`=(1.44xx10^(-4)mol^(2)L^(-2))/(4.8xx10^(-2)"mol"L^(-1))`
`=(14.4)/(4.8)xx10^(-3)"mol" L^(-1)`
`=3.3xx10^(-3)"mol" L^(-1)`