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For the reaction, N(2)O(4)(g) hArr 2NO(2...

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?

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The correct Answer is:
A, C
`K=([NO_(2)]^(2))/([N_(2)O_(4)])`
`=((1.61xx10^(-2))^(2)("mol" L^(-1))^(2))/(4.50xx10^(-2) "mol" L^(-1))`
`=5.67xx10^(-3) "mol" L^(-1)`
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