Answer:
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
