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For an equilibrium reaction, N(2)O(4)(g)...

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

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
`K_(C)=([NO_(2)]^(2))/([N_(2)O_(4)])=((1.2 xx 10^(-2))^(2))/(4.8 xx 10^(-2))`
`=3 xx 10^(-3)" mol L"^(-1)`
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NARAYNA-CHEMICAL EQUILIBRIUM-Exercise -II (H.W.)
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  15. In the dissociation of N(2)O(4) into NO(2), (1+ alpha) values with the...

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