For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) Suppose that 2.010 g of PCl5 is placed in an evacuated 555 mL bulb, which is then heated to 600. K. (a) What would be the pressure of PCl5 if it did not dissociate? WebAssign will check your answer for the correct number of significant figures. atm (b) What is the partial pressure of PCl5 at equilibrium? WebAssign will check your answer for the correct number of significant figures. atm (c) What is the total pressure in the bulb at equilibrium? WebAssign will check your answer for the correct number of significant figures. atm (d) What is the degree of dissociation of PCl5 at equilibrium?

a) pPCl5 = 0.856 atm

b)pPCl5 = 0.0557 atm

pCl2 = pCl3 = 0.800 atm

c)  Ptotal = 1.66 atm

d) 93.5

Explanation:

Step 1: Data given

Temperature = 600 K

Kp = 11.5

Mass of PCl5 = 2.010 grams

Volume of the bulb = 555 mL = 0.555 L

The bulb is heated to 600 K

Step 2: The balanced equation

PCl5(g) ⇄ PCl3(g) + Cl2(g)

Step 3:

a) pv = nrt

⇒with p = the pressure = TO BE DETERMINED

⇒with V = the volume = 0.555 L

⇒ with n =the number of moles PCl5 = 2.010 grams / 208.24 g/mol = 0.00965 moles

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

⇒ with T = the temperature = 600K

p = (0.00965 *0.08206*600)/0.555

pPCl5 = 0.856 atm

b)

The initial pressures

pPCl5 = 0.856 atm

pCl2 = pCl3 = 0 atm

For 1 mol PCl5 we'll have PCl3 and 1 mol Cl2

The pressure at the equilibrium

pPCl5 = (0.856 -x) atm

pCl2 = pCl3 = x atm

Kp = pPCl3 * pCl2/pPCl5  

11.5 = x*x / (0.856 - x)

11.5 = x²/(0.856- x)

x = 0.8003

pPCl5 = (0.856 -x) atm = 0.0557 atm

pCl2 = pCl3 = x atm = 0.800 atm

c) Since x = 0.8003 and PCl3 and PCl2 are x  

Ptotal = 0.8003 + 0.8003 +0.0557 = 1.66 atm

d)

The degree of dissociation = (x / initial pressure PCl5)

(0.8003/0.856) * 100 = 93.5

i think its 5049.

i hope this !