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An equilibrium mixture for the reaction,...

An equilibrium mixture for the reaction,
`2H_(2)S_((g)) Leftrightarrow 2H_(2(g))+S_(2(g))` had 1 mole of `H_(2)S`, 0.2 mole of `H_(2)` and 0.8 mole of `S_2` in a 2 litre flask. The value of `K_c` in mol `L^(-1)` is

A

0.16

B

0.016

C

0.08

D

0.004

Text Solution

Verified by Experts

The correct Answer is:
B
At eq, conc `("mol L"^(-1)) underset(1/2) (2H_(2)S) Leftrightarrow underset((0.2)/2) (2H_(2))+underset((0.8)/2)(S_(2))`
`K_(c)=([H_(2)]^2 [S_2])/([H_(2)S])^2=((0.2)/(2))^2 (0.8/2)/(1/2)^2=0.016`
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