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A mixtue of `H_(2) and I_(2) ` (vapour) in molecular proportion of 2: 3 was heated at ` 449^(@)C` till the reaction `H_(2) + I_(2) hArr 2 HI` reached equilibrium state . Calculate the percentage of iodine converted into `HI (K_(c) "at" 440^(@)C " is " 0*02).` |
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Answer» ` {:(,H_(2)(g),+,I_(2)(g),hArr,2HI(g)),(" Intial",2,,3,,0 "moles"),("Molar concs.",(2-x)/V,,(3-x)/V,,(2x)/V),(,,,,,):}` ` K_(c) = (2x//V)^(2)/([(2-x)//V][3-x)//V]=(4x^(2))/((2-x)(3-x))=0*02 " "` (Given) This on solving gives `x = 0*1615` ` :. % " of iodine converted into HI" = (0*1615)/3 xx 100 = 5* 38%` |
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