Question

1. consider the follwoing reaction:

Xe(g) +F2(g)------>XeF4(g)

A reaction mixture at 25 C initiallu contains 2.24 atm and 4.27 atm F2. If the equilibrium pressure of Xe is 0.34 atm,

a) find the equiibrium constant (Kp) for the reaction.

b) what is the value of (Kc) for this reaction.

please show all work

Answer 1

Xe(g) +F2(g)------>XeF4(g)

initially P_Xe   = 2.24 atm

P_F2   = 4.27 atm

at equilibrium = P_Xe   = 0.34 atm

means 2.24 - 0.34 = 1.9 atm consumed

so 1.9 atm F2 will consume

at equilibrium P_F2   = 4.27 - 1.9 = 2.37 atm

P_XeF4   = 1.9 atm

Kp = (P_XeF4)  / (P_Xe) (P_F2)   = (1.9) / 0.34 x 2.37

Kp = 1.9 / 0.8058

Kp = 2.34

b) Kp = Kc (RT)ⁿ

n = 1 - 2 = -1

temperature is not given by you so assuming 298 K

2.34 = Kc (0.0821 x 298)^-1

2.34 = Kc x 0.0409

Kc = 57.2