Question

A mixture of isobutylene (0.350 atm partial pressure at 500 K) and HCl(0.550 atm partial pressure at 500 K) is allowed to reach equilibrium at 500 K. What are the equilibrium partial pressures of tert-butyl chloride, isobutylene, and HCl?

The equilibrium constant KpKp for the gas-phase thermal decomposition of tert-butyl chloride is 3.45 at 500 K:
(CH3)3CCl(g)⇌(CH3)2C=CH2(g)+HCl(g)

Answer 1

For the reaction:  
(CH₃)₃CCl(g) ⇌ (CH₃)₂C=CH₂(g) + HCl(g)

K_p = 3.45 at 500K

K_p for this reaction can be written as:

Here P_x is the partial pressure at equilbrium of each species.

Now, initial partial pressures are: and P_HCl = 0.550 atm

So, the above reaction will move backwards (as partial pressure of (CH₃)₃CCl(g) is 0atm initially)

Now, constructing reaction pressure table at equilbrium after time 't' when x amount of partial pressures of (CH₃)₂C=CH₂(g) and HCl(g) has been reduced to form (CH₃)₃CCl(g) :

	(CH3)3CCl(g)	⇌	(CH3)2C=CH2(g)	+ HCl(g)
Initial partial pressure	0		0.350	0.550
Change in partial pressure	+x		-x	-x
Final partial pressure at equilibrium	x		(0.350-x)	(0.550-x)

As we know:

So,

Solving this we get:

Solving this quadratic equation:
x = 4.30 or x = 0.045

As we know x must be lesser than 0.350 and 0.550 as equilbrium pressures can not be negative. So, we can reject x = 4.30
Hence, we get: x =0.045

Equilbrium partial pressures:
(CH₃)₃CCl(g) : x = 0.045 atm

(CH₃)₂C=CH₂(g) : (0.350-x)= (0.350 - 0.045) = 0.305 atm

HCl(g) : (0.550-x) = (0.550- 0.045) = 0.505 atm