1.

An equlibrium mixture of the reaction 2H_(2)S_((g))hArr2H_(2(g))+S_(2(g))"had 0.5 mole"H_(2)S, 0.10 "mole" H_(2)and 0.4 "mole" S_(2) in one litre vessel. The vlaue of equlibrium constant (K) in mole "litre"^(-1) is

Answer»

`0.004`
`0.008`
`0.016`
`0.160`

SOLUTION :`K=([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2))=([0.10]^(2)[0.4])/([0.5]^(2))=0.016`


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