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For the hypothetical reaction A +B hArr C+2D equilibrium constant at 400 K is 1.8 xx 10^(-6) mol L^(-1) |
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Answer» `9.5 xx 10^(-5) mol L^(-1)` 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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