Question

For the following reaction at 600. K, the equilibrium constant,Kp, is 11.5.PCl5(g) reverse reaction arrowPCl3(g) + Cl2(g)Suppose that 2.070 g ofPCl5 is placed in an evacuated 505 mL bulb, which is then heated to 600. K.(a) What would be the pressure ofPCl5 if it did not dissociate?1 atm(b) What is the partial pressure of PCl5 atequilibrium?2 atm(c) What is the total pressure in the bulb at equilibrium?3 atm(d) What is the degree of dissociation of PCl5 atequilibrium?4%

Answer 1

a) What would be the pressure of PCl5 if it did not dissociate? = 0.965 atm

b) What is the partial pressure of PCl5 atequilibrium? = 0.0697 atm

c) What is the total pressure in the bulb at equilibrium? = 1.86 atm

d) What is the degree of dissociation of PCl5 at the equilibrium? = 92.8%

Step-by-step explanation:

Step 1: Data given

Temperature = 600 K

Kp = 11.5

Mass of PCl5 = 2.070 grams

Volume = 505 mL = 0.505 L

Step 2: The balanced equation

PCl5(g) → PCl3(g) + Cl2(g)

Step 3: What would be the pressure of PCl5 if it did not dissociate?

Moles PCl5 = mass PCl5 / Molar mass PCl5

Moles PCl5 = 2.070 grams / 208.24 g/mol

Moles PCl5 = 0.0099 moles

pV = nRT

⇒with p = the pressure of the gas = TO BE DETERMINED

⇒with V = the volume = 505 mL = 0.505 L

⇒with n = the number of moles =0.0099 moles

⇒with R = the gas constant = 0.08206 L*atm/K*mol

⇒with T = the temperature = 600 Kelvin

p = nRT/V

p = (0.0099*0.08206*600)/0.505

p =0.965 atm

b) What is the partial pressure of PCl5 atequilibrium?

Kp = (pPCl3) * (pCl2)/(pPCl5 )

11.5 = (x)(x)/(0.965atm - x)

11.5 = x²/(0.965atm - x)

x2 + 11.5x - 11.097 = 0

x = 0.8953

pPCl3 = 0.8953 atm = pCl2

pPCl5 = 0.965atm - 0.8953 atm = 0.0697 atm

At equilibrium PCl5 partial pressure is 0.0697 atm

c) What is the total pressure in the bulb at equilibrium?

Since x = 0.8953 atm and PCl3 = PCl2 = x atm

Ptot = 0.0697 + 0.8953 +0.8953 = 1.86atm

d) What is the degree of dissociation of PCl5 at the equilibrium?

(0.8953atm/0.965 atm) * 100% = 92.8%