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%

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%

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%

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