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MAXImum [283]
3 years ago
5

Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibri

um 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.)
Chemistry
1 answer:
melisa1 [442]3 years ago
5 0

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

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