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3.00 mol of PCl(5) kept in 1 L closed re...

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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` {:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Initial",3*00 " mol ",,0,,0),("At. eqm.",(3*0-x)"mol",,x " mol",,x" mol "),("Molar conc",(3*0-x),,x,,x):}` (as volume of vessel = 1 L)
` K_(c) = ([PC_(3)] [Cl_(2)])/([PC_(5)])`
` 1*80 = (x xxx)/(3-x)=x^(2)/(3-x)or x^(2) = 5*40 - 1*80 x or x^(2) + 1*80 x - 5* 40=0`
` :. x= (-bpm sqrt(b^(2) - 4ac))/(2a)=(-1*80 pm 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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