JEE Class main Answered

At temp T a co mpound AB2(g) dissociates according to the reaction 2AB2→2AB+B2 with a degree of dissociation X which is small as compared to unity .the expression for Kpin terms of X and total pressure P is

Asked by ak020820002 | 05 Feb, 2019, 00:48: AM

Expert Answer

The compound AB_2(g) dissociates according to the reaction,

$space space space space space space space space space space space space space space space 2 AB subscript 2 space space end subscript space space rightwards arrow space space space 2 AB space space plus space space straight B subscript 2 space initial space space space space space space space space space space space space 1 space space space space space space space space space space space space space space 0 space space space space space space space space space 0 At space equ space space space space space space 2 open parentheses 1 minus straight x close parentheses space space space space space space space space 2 straight x space space space space space space space space straight x where comma straight x space equals space degree space of space dissociation Total space moles space at space equ space equals space space space space 2 open parentheses 1 minus straight x close parentheses space plus space space 2 straight x space space plus space space straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 minus 2 straight x plus space 2 straight x plus space straight x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 2 plus straight x close parentheses straight P subscript AB space equals space fraction numerator 2 open parentheses 1 minus straight x close parentheses straight P over denominator open parentheses 2 plus straight x close parentheses end fraction straight P subscript AB space equals space fraction numerator 2 xP over denominator open parentheses 2 plus straight x close parentheses end fraction straight P subscript straight B subscript 2 end subscript space space equals space fraction numerator xP over denominator open parentheses 2 plus straight x close parentheses end fraction space straight K subscript straight P space space space space equals space space fraction numerator open parentheses begin display style fraction numerator 2 xP over denominator 2 plus straight x end fraction end style close parentheses squared space open parentheses begin display style fraction numerator straight x over denominator 2 plus straight x end fraction straight P end style close parentheses over denominator open parentheses begin display style fraction numerator 2 open parentheses 1 minus straight x close parentheses over denominator 2 plus straight x end fraction straight P end style close parentheses end fraction space space space space space space space space equals fraction numerator straight x cubed straight P over denominator open parentheses 2 plus straight x close parentheses open parentheses 1 minus straight x close parentheses squared end fraction As space it space is space given space that space straight x less than less than 1 space therefore open parentheses 1 minus straight x close parentheses almost equal to space 1 space and space open parentheses 2 plus straight x close parentheses almost equal to space 2 space space straight K subscript straight P space space equals space space open parentheses fraction numerator straight x cubed straight P over denominator 2 end fraction close parentheses$