Question

Consider the following gas phase dissociation, PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g) with equilibrium constant K$_p$ at a particular temperature and at pressure P. The degree of dissociation ($\alpha$) for PCl$_5$(g) is
PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g)

• A. $\alpha = \left(\frac{K_p}{K_p+P}\right)^{1/3}$
• B. $\alpha = \left(\frac{K_p}{K_p+P}\right)$
• C. $\alpha = \left(\frac{K_p}{K_p+P}\right)^{1/2}$
• D. $\alpha = \left(\frac{K_p}{K_p+P}\right)^{2}$

Hint:

Using a simple frame or just bolding for the box
Key Points:
Use an ICE table with the degree of dissociation ($\alpha$).
Calculate total moles at equilibrium to find mole fractions.
Partial pressure = Mole Fraction $\times$ Total Pressure (P).
Write the K$_p$ expression in terms of partial pressures.
Substitute and solve for $\alpha$.
Remember the difference of squares: $(1+\alpha)(1-\alpha) = 1-\alpha^2$.

Answer 1

The Correct Option is C

Let the initial moles of PCl₅ be 1 and the degree of dissociation be α. We can set up an ICE table based on moles:

Reaction:	PCl5(g)		⇌		PCl3(g)	+	Cl2(g)
Initial (mol):	1				0		0
Change (mol):	-α				+α		+α
Equil. (mol):	1-α				α		α

Total moles at equilibrium = (1-α) + α + α = 1+α. 
The total pressure at equilibrium is given as P. Now, calculate the partial pressures using p_i = X_i × P_total, where X_i is the mole fraction. p_PCl5 = ( (1-α) / (1+α) ) P 
p_PCl3 = ( α / (1+α) ) P 
p_Cl2 = ( α / (1+α) ) P The expression for the equilibrium constant K_p is: 
K_p = (p_PCl3 × p_Cl2) / p_PCl5 Substitute the partial pressures: 
K_p = (( αP / (1+α) ) ( αP / (1+α) )) / (( (1-α)P ) / (1+α) ) K_p = (α²P² / (1+α)²) / (( (1-α)P ) / (1+α) ) K_p 
= (α²P² / (1+α)²) × ((1+α) / ((1-α)P)) K_p = (α²P) / ((1+α)(1-α)) K_p = (α²P) / (1-α²) 
Now, rearrange to solve for α: 
K_p (1-α²) = α² P K_p - K_p α² = α² P K_p 
= α² P + K_p α² K_p = α² (P + K_p) α² 
= K_p / (P + K_p) α = √( K_p / (K_p + P) ) = ( K_p / (K_p + P) )^1/2 
This corresponds to option (C).