For a reaction, `H_2+I_2hArr 2HI` at 721 K , the value of equilibrium constant is 50. If 0.5 moles each of `H_2` and `I_2` is added to the system the value of equilibrium constant will be :

Options

A. 0.02
B. 0.2
C. 50
D. 25

Answer

C

Explanation

`K=([HI]^(2))/([H_(2)][I_(2)])=50=(4alpha^(2))/((1-alpha)^(2))`
`=sqrt(50)=(2a)/(1-a)=7.07`
or `2alpha=7.07-7.07 alpha`
`9.07alpha=7.07 ` or `alpha=(7.07)/(9.07)=0.78`
If 0.5 moles each of `H_(2)` and `I_(2)` are taken than Equilibrium conc. `[H_(2)]=0.5-0.39, `
`[I_(2)]=0.5-0.39,HI=0.78`
`K=(0.78xx0.78)/(0.11xx0.11)=50`