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InterviewSolution
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3.00 mol of `PCl_(5)` kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate the composition of the mixture at equilibrium. `K_(c) = 1.80.` |
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Answer» `{:(,PCl_(5)(g) ,hArr, PCl_(3)(g) ,+,Cl_(2)(g)),("Initial molar conc.",3.00,,0,,0),("Eqm. molar conc." ,(3.0-x),,x,,x):}` `K_(c)=[[PCl_(3)][Cl_(2)]]/[[PCl_(5)]] or 1.80 =(x xx x )/(3-x) = (x^(2))/(3-x) or x^(2) = 5.40 -1.80 x ror x^(2) +1.80 x -5.40 =0` `:. x= (-bunderset(-)(+)sqrt(b^(2)-4ac))/(2a) =(-1.80 underset(-)(+)sqrt((1.80)^(2)-4(-5.40)))/(2) = 1.59 ("neglecting -ve value")` `:. " At . equilibrium " ,[PCl_(5)] = 3 - 1.59 = 1.41 M , [PCl_(3)] =[Cl_(2)] = 1.59 M` |
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