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A flask is filled with PCl5 to a pressure of 2.00 atm at 300°C and allowed to come to equilibrium according to the reaction: PCl5(g) ⇄ PCl3(g) + Cl2(g) Analysis shows the total pressure in the flask at equilibrium is 3.96 atm. Calculate the equilibrium constant Kp for the reaction.

Respuesta :

aimartinezm

Answer:

96.04

Explanation:

Kp is defined for any reaction as:

aA(g) +bB(g) ⇌ cC(g) +dD(g)

[tex]Kp= \frac{partial-pressure-of-products}{partial-pressure-of-reagents}[/tex]

[tex]Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}[/tex]

For the reaction:

[tex] PCl_{5}(g) equilibrium PCl_{3}(g) + Cl_{2}(g)[/tex]

Kp is:

[tex]Kp=\frac{P_{PCl_{3}}*P_{Cl_{2}}}{P_{PCl_{5}}}[/tex]

P in the equation is the partial pressure for each specie in the equilibrium.

We know the total pressure for the system at the beginning is 2 atm. We can say that at the beginning we have just [tex]PCl_{5}[/tex] so the total pressure for the system is also the partial pressure for [tex]PCl_{5}[/tex]  

In the equilibrium we have the three species [tex] PCl_{5}[/tex],[tex]PCl_{3}[/tex] and [tex]Cl_{2}[/tex] so now we need to estimate the partial pressure for each specie.  

We know that the total pressure for the system in the equilibrium is 3.96 atm. The total pressure is equal to the sum of the partial pressure of each component in the mix as:

[tex]P_{total}=P_ {PCl_{3}}+ P_{Cl_{2}}+ P_{PCl_{5}} = 3.96 atm[/tex]  

Until we reach the equilibrium [tex] PCl_{5}[/tex] it’s been used to produce [tex] PCl_{3}[/tex] and [tex]Cl_{2}[/tex]. We can call (x) the among of  [tex] PCl_{5}[/tex] that reacted until the system reach the equilibrium so [tex] P_{PCl_{5}}[/tex] in the equilibrium will be

[tex](P_{PCl_{5}}^{initial}) -  x=P_{PCl_{5}}^{equilibrium}[/tex] 

When reviewing the molar ratio in the reaction we see that it is 1: 1 so the amount of [tex] PCl_{5}[/tex] that reacted will be equal to the amount of [tex] PCl_{3}[/tex] and [tex]Cl_{2}[/tex] produced.

                           PCl5      Cl2                PCl3

Initial             2 atm          0 atm       0 atm  

Equilibrium      (2 atm - x) X atm       X atm

Using the equation for total pressure:

[tex]P_{total}= (2atm-x)+x+x = 3.96 atm[/tex] 

solving we get that x = 1.96 atm.

This is the partial pressure for [tex] PCl_{3}[/tex] and [tex]Cl_{2}[/tex] in the equilibrium  

For each component we have:

[tex]P_{PCl_{5}}=2 atm - 1.96 atm = 0.04 atm[/tex]

[tex]P_{PCl_{3}}= 1.96 atm[/tex]

[tex]P_{Cl_{2}}= 1.96 atm[/tex]

So

[tex]Kp=\frac{1.96 atm*1.96 atm}{0.04 atm}=96.04[/tex]