4.5 moles each of hydrogen and iodine heated in a seated in a sealed ten litre vessel. At equlibrium, 3 moles of HI were found. The equilibrium constnt for `H_(2(g))hArrPCl_(3(g))+Cl_(2(g))` is

Options

A. 1
B. 10
C. 5
D. `0.33`

Answer

A

Explanation

`H_(2)+I_(2)hArr2HI`
`{:("Initial conc.",4.5,4.5,0),("at Equlibrium",4.5-x,4.5-x,2x):}`
From euestion `2x=3`
`x=3/2=1.5`
So conc. At eqm. `4.5-1.5of H_(2)`
`=4.5-1.5of I_(2)and 3 of HI`
`K=([H]^(2))/([I_(2)][H_(2)])=(3xx3)/(3xx3)=1.`