Question

Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibrium mixture in a 5.00 L flask at 245°C contains 4.13 g of PCl5, 8.87 g of PCl3, and 2.90 g of Cl2. How many grams of each will be found if the mixture is transferred into a 2.00 L flask at the same temperature? (Enter unrounded values.)

Answer 1

Answer:

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

Explanation:

The equilibrium constant will remain the same irrespective of the amount of reactant taken.

Let us calculate the equilibrium constant of the reaction.

Kc=$\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}$

Let us calculate the moles of each present at equilibrium

$moles=\frac{mass}{molarmass}$

molar mass of PCl₅=208

molar mass of PCl₃=137

molar mass of Cl₂=71

moles of PCl₅ = $\frac{mass}{molarmass}=\frac{4.13}{208}=0.02$

moles of PCl₃= $\frac{mass}{molarmass}=\frac{8.87}{137}=0.06$

moles of Cl₂ = $\frac{mass}{molarmass}=\frac{2.90}{71}=0.04$

the volume is 5 L

So concentration will be moles per unit volume

Putting values

Kc = $\frac{\frac{0.06}{5}\frac{0.04}{5}}{\frac{0.02}{5}}=0.024$

Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow

                $PCl_{5}--->PCl_{3}+Cl_{2}$

Initial                 0.02           0.06       0.04

Change             -x                   +x          +x

Equilibrium     0.02-x           0.06+x    0.04+x

Conc.                (0.02-x)/2       (0.06+x)/2   (0.04+x)/2

Putting values

0.024 = $\frac{(0.06+x)(0.04+x)}{(0.02-x)2}$

Solving

$(0.024(2)(0.02-x)=(0.06+x)(0.04+x)$

$0.00096-0.048x=0.0024+x^{2}+0.1x$

$0.148x+x^{2}+0.00144=0$

x = -0.01

so the new moles of

PCl₅ = 0.02 + 0.01  =0.03

PCl₃ = 0.06-0.01 = 0.05

Cl₂ = 0.04-0.01 = 0.03

mass of each will be:

mass= moles X molar mass

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g