Question

On a give condition, the equilibrium concentration of `HI, H_(2) and I_(2)` are `0.80,1.10` mole/litre. The equlibrium constant for the reaction `H_(2)+I_(2)hArr 2HI` will be
A. 64
B. 12
C. 8
D. `0.8`

Answer 1

Correct Answer - A
`H_(2)+I_(2)hArr2HI,[HI]=0.08,[H_(2)]=0.10,[I_(2)]=0.10`
`K_(c)=([HI]^(2))/([H_(2)][I_(2)])=(0.08xx0.80)/(0.10xx0.10)=64`