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For a reaction X+Y = 2Z , 1.0 mo of X 1....

For a reaction X+Y = 2Z , 1.0 mo of X 1.5 mol of Y and 0.5 mol of Z where taken in a 1L vessel and allowed to react . At equilibrium , the concentration of Z was `1.0 mol L^(-1)` . The equilibrium constant of the reaction is _____ ` (x)/(15) ` . The value of x is _____.

Text Solution

Verified by Experts

The correct Answer is:
16
`{:(,X +,Y hArr, 2Z),(t= 0,1 mol,1.5 mol,0.5 mol):}`
Since moles of Z are increased at equilibrium therefore reaction goes in forward direction to attain the equilibrium
`{:(,x +,y hArr ,2Z),(t = t_(eq),1-a,1.5 -a,0.5 +2a = 1 "mole"),(,,,rArr a = 0.25 ),(x +,t to,2Z,),(0.75 mol,1.25 mol,1 mol,):}`
`K_(eq) = ([Z]^(2))/([X][Y])= (1)/(0.75 xx [1.25])=(X)/(15)`
`X = (15)/((0.75 xx 1.25)) = 16`
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