Answer to Question #195440 in Inorganic Chemistry for Lovely Mae Zolina

Question

Phosphorus pentachloride (Kp=0.0870) is one of the most important phosphorus chlorides, and is use as

a chlorinating reagent. It is produced by reacting chlorine gas and Phosphorus trichloride (PCl3) at

300°C. PCl3(g)+Cl2(g)<-> PCl5(g). In an industrial set-up, the stainless steel container was charged

with 020 atm PCl3, 0.20 atm Cl2, and 0.50 atm PCl5 at this temperature. (a) Use the reaction quotient

to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium

partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole

fraction of Cl2 in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing

the temperature of the system have on the mole fraction of Cl2 in the equilibrium mixture

Answer 1

a) Q =reaction quotient

k=k_p= equilibrium constant

$Q ={ [PCl5 ]\over [PCl3 ][Cl2 ]}$ = $Q ={ [0.50 ]\over [0.20 ][0.20 ]}=12.5$

as Q>k reaction will go backward from right to left.

b) for this we will use variable x.

as

at the equilibrium point .

PCl₃(g)+Cl₂(g)<-> PCl₅(g)

partial [ressure at equi. point

PCl₃(g)=0.20-x

Cl₂(g)=0.20-x

PCl₅(g)=0.50+x

and putting values in

$k ={ [PCl5 ]\over [PCl3 ][Cl2 ]}$

$0.0870 ={ [0.50+x ]\over [0.20-x ][0.20-x ]}$

we get x= -0.462

so partial pressures at equilibrium are

PCl₃(g)=0.20- (-0.462)=0.662

Cl₂(g)=0.20-( -0.462)=0.662

PCl₅(g)=0.50+ (-0.462)=0.038

c) according to le-chateliar's principle

if $\Delta$ ng <0 and increasing volume

then equn shift in backword direction

thus amount of Cl₂ will increase.

and Cl2 increase then mole fraction will increase.

d) given exothermic $\Delta$H<0 then temp. increase will shift equilibrium backword. then amount of Cl₂ will increase.

and Cl2 increase then mole fraction will increase.