Question

1) Consider the following reaction: Xe(g) + 2 F2(g) ⇌ XeF4(g)

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A reaction mixture initially contains 3.24 atm Xe and 5.27 atm F2. If Kp = 4.1 x 10 , what is the

equilibrium concentration for XeF4?

2) Antimony trifluoride and Antimony pentafluoride equilibrate in the presence of molecular fluorine according to the reaction:

SbF5(g) ⇌ SbF3(g) + F2(g)

An equilibrium mixture at 250 K contains

SbF3 = 0.104 atm,

F2 = 0.257 atm, and

SbF5 = 4.3 atm. What is the value of Kp at this temperature?

Answer 1

For any equilibrium reaction

aA + bB ⇌ cC + dD

value of the equilibrium constant, K_P = (P_C)^c * (P_D)^d / (P_A )^a * (P_B)^b   

here, in the given question the reaction is

1)  Xe(g) + 2 F2(g)  ⇌ XeF4(g)

Initial pressure of Xe = 3.24 atm

Initial pressure of F2 = 5.27 atm

Intial pressure of XeF4 = 0 atm (because reaction is yet not started)

Lets assume at equilibrium if x atm pressure is decrease from Xe and 2x atm from F2 and there is an increase of x atm pressure for XeF4

So,

Equilibrium pressure for Xe = 3.24 - x atm

Equilibrium presssure for F2 = 5.27 - 2x atm

Equilibrium pressure for XeF4 = x atm

So, on applying the formula for equilibrium constant ,

Kp = x / (3.24-x)  * (5.27 - 2x)²

after putting the value of Kp = 4.1 * 10^-5

we will get x = 0.00367 atm and x denotes the equilibrium concentration for XeF4

2) SbF5(g)  ⇌ SbF3(g) + F2(g)

Equilibrium pressure of SbF5 = 4.3 atm

Equilibrium pressure of SbF3 = 0.104 atm

Equilibrium pressure of F2 = 0.257 atm

after applying the formula of equilibrium constant,

K_p = (0.104)*(0.257) / (4.3)

K_p = 0.006215 = 6.215 * 10^-3 (This is the required answer)