Answer:
PClâ‚… = 223.4 torr
PCl₃ = 6.8 torr
Clâ‚‚ = 26.4 torr
Explanation:
For gas substances, the equilibrium constant can be calcultaed based on the partial pressures (Kp). For a generic reaction:
aA(g) + bB(g) ⇄ cC(g) + dD(g)
[tex]Kp = \frac{(pC)^c*(pD)^d}{(pA)^a*(pB)^b}[/tex], where pX is the partial pressure of X.
The reaction with the gas mixture given is:
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Kp = [(pPCl₃)*(pCl₂)]/(pPCl₅)
Kp = [13.2*13.2]/217
Kp = 0.803
When more Cl₂ is added, for Le Chatêlier's principle, the equilibrium will shift for the left, more PCl₅ will be formed, and the equilibrium will be reestablished.
The initial total pressure was 243.4 torr, so if it jumps to 263.0 torr, it was added 19.6 torr of Clâ‚‚, so the partial pressure of Clâ‚‚ is 32.8 torr. For the reaction:
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
217.0 Â Â Â Â 13.2 Â Â Â Â 32.8 Â Â Â Initial
+x        -x       -x     Reacts (stoichiometry is 1:1:1)
217 + x    13.2-x   32.8-x  Equilibrium
So, the equilibrium constant must be:
[tex]Kp = \frac{(13.2-x)*(32.8-x)}{217+x}[/tex]
0.803 = (432.96 - 46x + x²)/(217 + x)
432.96 - 46x + x² = 174.251 + 0.803x
x² - 46.803x + 258.71 = 0
By Bhaskara's equation:
Δ = (46.803)² - 4*1*258.71
Δ = 1,155.68
x =[-(- 46.803) ±√1,155.68]/2
x' = (46.803 + 33.99)/2
x' = 40.40
x'' = (46.803 - 33.99)/2
x'' = 6.40
x < 32.8, so x = 6.40
The new partial pressures are:
PClâ‚… = 217.0 + 6.40 = 223.4 torr
PCl₃ = 13.2 - 6.40 = 6.8 torr
Clâ‚‚ = 32.8 - 6.40 = 26.4 torr
