For the equilibrium Br_2(g) + CI_2(g) 2 BrCI(g), the equilibrium constant K_p is
ID: 953384 • Letter: F
Question
For the equilibrium Br_2(g) + CI_2(g) 2 BrCI(g), the equilibrium constant K_p is 7.0 at 400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCI? (a) 0.57 atm. (b) 0.22 atm. (c) 0.45 atm. (d) 0.15 atm. (c) 0.31 atm. For the equilibrium PCI_5 PCI_3(g) + CI_2(g), the equilibrium constant K_p is 0.497 at 500 K. A gas cylinder at 500 K is charged with PCI_5(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCI_5, PCI_5, and CI_2 at this temperature?Explanation / Answer
1 )
Br2 + Cl2 ----------------------------> 2 BrCl
0 -------0 ------------------------------- 1 ------------------> initial
x x 1-2x ---------------> equilibrium
Kp = PBrCl^2 / PBr2 x PCl2
7.0 = (1-2x)^2 / x^2
7 x^2 = 1 + 4x^2 -4x
3x^2 +4x -1 = 0
x = 0.215
equilibrium pressure of BrCl = 1- 2x
= 1 - 2x 0.215
= 0.57 atm
answer : a ) 0.57 atm
2)
PCl5 <-----------------------> PCl3 + Cl2
1.66 0 0 -------------> initial
1.66-x x x --------------> equilibrium
Kp = x^2 / 1.66-x
0.497 = x^2 / 1.66-x
x^2 + 0.497 x - 0.825 = 0
x = 0.693
equilibrium pressures :
PCl5 pressure = 1.66 - x = 1.66 -0.693 = 0.967 atm
Cl2 pressure = x = 0.693 atm
PCl3 pressure = x = 0.693 atm