Q.

2 moles each of A and B were taken in a flask at some temperature so as to establish the following equilibrium A+B2D It was observed that 13 of A has reacted. After attainment of equilibria 2 mole of C was added which caused establishment of another equilibria B+CE+D If at new equilibria mole fraction of E is 16, then calculate % of A dissociated after both equilibria has established. Given: 61=7.81. Give answer after multiplying by 100. 

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answer is 675.

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Detailed Solution

In the reaction of A and B;

A           +      B              2D223 223      43

KC1=1

Final moles of E = 1., as Δng=0 for both reactions

 At new equilibrium

A        +     B               2D(2x) (2xy) (2x+y)B            +        C          E+D(2xy) (2y)        y (2x+y)

Clearly, y = 1

=> In 1 st equilibrium,

(2x+1)2(1x)(2x)=13x2+7x1=0x=7±616 % diss. of A=0.8162×100=6.75%

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