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For the reaction equilibrium, N2O(4(g)) ...

For the reaction equilibrium, `N_2O_(4(g)) Leftrightarrow 2NO_(2(g))` the concentrations of `N_2O_4 and NO_2` at equilibrium are `4.8 xx 10^(-2) and 1.2 xx 10^(-2)" mol L"^(-1)` respectively. The value of `K_c` for the reaction is

A

`3.3 xx 10^(2)" mol L"^(-1)`

B

`3.3 xx 10^(-1)" mol L"^(-1)`

C

`3.3 xx 10^(-3)" mol L"^(-1)`

D

`3.3 xx 10^(3)" mol L"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`N_2O_4 to 2NO_2`
`K_(c)=([NO_2]^2)/([N_2O_4])=((1.2 xx 10^(-2))^2)/(4.8 xx 10^(-2))=3.0 xx 10^(-3)`
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