Question

At `440^(@)C`, the equilibrium constant (K) for the following reaction is `49.5, H_(2)(g)+I_(2)(g) hArr 2HI(g)`. If `0.2` mol of `H_(2)` and `0.2` mol of `I_(2)` are placed in a `10-L` vessel and permitted to react at this temperature, what will be the concentration of each substance at equilibrium?

Answer 1

Correct Answer - A::B::C::D
`{:(,H_(2),+,I_(2),rarr,HI),("Initial",0.2,,0.2,,),("At. equilibrium",(0.2-x)/(10),,(0.2-x)/(10),,(2x)/(10)):}`
`49.5=K=([HI]^(2))/([H_(2)][I_(2)]), x=0.1557`
`[H_(2)]=[I_(2)]_(eq.)=(0.2-0.1557)/(10)=0.00443 M`
`[HI]=(2xx0.1557)/(10)=0.03114 M`