1.

For the hypothetical reaction `A +B hArr C+2D` equilibrium constant at 400 K is `1.8 xx 10^(-6)` mol `L^(-1)`A. `9.5 xx 10^(-5) mol L^(-1)`B. `9.5 xx 10^(-4) mol L^(-1)`C. `4.75 xx10^(-4) mol L^(-1)`D. None of these

Answer» Correct Answer - B
`{:(,A,+,B,hArr,C,+,2D),("Initial conc." (mol L^(-1)),-,,-,,1.0,,1.0),("At equilibrium" (mol L^(-1)),((1-0-x)/(2)),,((1.0-x)/(2)),,((1.0+x)/(2)),,x):}`
D used up `=(1.0 -x )` mol `L^(-1)`
C used up `=((1.0-x)/(2)) mol L^(-1)`
C at equilibrium `(1.0-(1.0-x)/(2))mol L^(-1)`
`=((1.0+x)/(2)) mol L^(-1)`
A ( or B ) produced at equilibrium `=C` used up
`=((1.0-x)/(2))mol L^(-1)`
A very small value of K indicates that at equilibrium concentrations of C and D are very small i.e., x is very small. As such at equilibrium
`[C]=((1.0+x)/(2))=0.5 mol L^(-1)`
`[D]=x mol L^(-1) , [A]=[B]`
`=((1.0-x)/(2)) mol L^(-1)`
`=0.50 mol L^(-1)`
`:. K=([C][D]^(2))/([A][B])`
`1.8 xx 10^(-6)=(0.50xx x^(2))/(0.50 xx 0.80)`
`x^(2)=9.0 xx 10^(-9)`
`implies x=9.5 xx10^(-4) mol L^(-1)`
`:. [D]=x=9.5 xx 10^(-4) mol L^(-1)`


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