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InterviewSolution
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The equilibrium constant for the reaction `A_(2)(g) +B_(2) hArr 2AB(g)` at 373 K is 50. If one litre flask containing one mole of `A_(2)` is connected to a two flask containing two moles of `B_(2)`, how many moles of AB will be formed at 373 K? |
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Answer» Total volume available for reaction mixture =3 L The value of equilibrium constant =50 Let x moles of both the reactants `A_(2)` and `B_(2)` react to form 2x moles of AB according to the equation given above. Therefore, the molar concentration per litre of the different species at equilibrium point is : `{:(,A_(2)(g),+,B_(2)(g),hArr,2AB(g)),("Initial moles//litre" ,1,,2,,0),("Moles //litre at equilibrium point" ,(1-x)/(3),,(2-x)/(3),,(2x)/(3)):}` Applying Law of chemical equilibrium `K_(c) =[[AB]^(2)]/[[A_(2)][B_(2)]]=((2x)/(3))^(2)/(((1-x)/(3))xx((2-x)/(3))^(2))` `50 = (4x^(2))/((1-x)(2-x))` `2x^(2) =25 (2-3x+x^(2)) , 23x^(2) -75x + 50 =0` By solving quadratic equation `.^(**) x= (75 underset(_)(+) sqrt((75^(2)-4 xx 23 xx50)))/(2xx23) =2.326 or 0.934` |
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