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At 400^(@)C,H(2)(g) and I(2)(g) are allo...

At `400^(@)C,H_(2)(g) and I_(2)(g)` are allowed to react in a closed vessel of 5 L capacity to produce H(g). At equilibrium, the mixture in the flask is found to consist of 0.6 mol `H_(2)(g),0.6" mol "I_(2)(g)` annd 3.5 mol `HI(g)`. Determine the value of `K_(c)` of the reaction.

Text Solution

Verified by Experts

Equilibrium of the reaction is: `H_(2)(g)+I_(2)(g)hArr2HI(g)`
Therefore, `K_(c)=([HI]^(2))/([H_(2)]xx[I_(2)])`
Capacity of the vessel=5L. Hence, molar concentrations of `H_(2),I_(2) and HI` are: `[H_(2)]=(0.6)/(5)=0.12mol*L^(-1)`
`[I_(2)]=(0.6)/(5)=0.12mol*L^(-1) and [HI]=(3.5)/(5)=0.7mol*L^(-1)`
`therefore K_(c)=((0.7)^(2))/((0.12)xx(0.12))=34.03`.
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