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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.)

Respuesta :

Answer:

PClâ‚… = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Clâ‚‚ = 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=[tex]\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}[/tex]

Let us calculate the moles of each present at equilibrium

[tex]moles=\frac{mass}{molarmass}[/tex]

molar mass of PClâ‚…=208

molar mass of PCl₃=137

molar mass of Clâ‚‚=71

moles of PClâ‚… = [tex]\frac{mass}{molarmass}=\frac{4.13}{208}=0.02[/tex]

moles of PCl₃= [tex]\frac{mass}{molarmass}=\frac{8.87}{137}=0.06[/tex]

moles of Clâ‚‚ = [tex]\frac{mass}{molarmass}=\frac{2.90}{71}=0.04[/tex]

the volume is 5 L

So concentration will be moles per unit volume

Putting values

Kc = [tex]\frac{\frac{0.06}{5}\frac{0.04}{5}}{\frac{0.02}{5}}=0.024[/tex]

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

                [tex]PCl_{5}--->PCl_{3}+Cl_{2}[/tex]

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 = [tex]\frac{(0.06+x)(0.04+x)}{(0.02-x)2}[/tex]

Solving

[tex](0.024(2)(0.02-x)=(0.06+x)(0.04+x)[/tex]

[tex]0.00096-0.048x=0.0024+x^{2}+0.1x[/tex]

[tex]0.148x+x^{2}+0.00144=0[/tex]

x = -0.01

so the new moles of

PCl₅ = 0.02 + 0.01  =0.03

PCl₃ = 0.06-0.01 = 0.05

Clâ‚‚ = 0.04-0.01 = 0.03

mass of each will be:

mass= moles X molar mass

PClâ‚… = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Clâ‚‚ = 0.03X71 = 2.13 g

okpalawalter8

The new mass of each compound is mathematically given as

PCl5= 6.24g

PCl3 =6.85 g

Cl2= 2.13 g

What grams of each compound will be given if the mixture is transferred into a 2.00 L flask at the same temperature?

Generally, the equation for the  equilibrium constant   is mathematically given as

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

Where

moles of PCl5 = 0.02

moles of PCl3=0.06

moles of Cl2=  0.04

Thereofore

[tex]Kc = \frac{\frac{0.06}{5}\frac{0.04}{5}}{\frac{0.02}{5}}\\\\Kc=0.024[/tex]

Generally, the equation for the  Chemical Reaction   is mathematically given as

PCl5--->PCl3+Cl2

Therefore

[tex]0.024 = \frac{(0.06+x)(0.04+x)}{(0.02-x)2}[/tex]

x=-0.01

Hence the resulting moles is

PCl5 = 0.02 + 0.01  

PCl5=0.03

PCl3 = 0.06-0.01

PCl3= 0.05

Cl2 = 0.04-0.01

Cl2= 0.03

In conclusion

mass= moles * molar mass

PCl5 = 0.03 * 208

PCl5= 6.24g

PCl3 = 0.05 * 137

PCl3 =6.85 g

Cl2 = 0.03*71

Cl2= 2.13 g

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