PCl_{5} dissociates as:

PCl_{5} ⇌ PCl_{3} Cl_{2}

If α is the degree of dissociation at certain temperature under atmospheric pressure, then

Initial concentration:

PCl_{5} = 1

PCl_{3} = 0

Cl_{2} = 0

At Equilibrium:

PCl_{5} = 1 – α

PCl_{3} = α

Cl_{2} = α

Total number of moles at equilibrium = 1 – α + α + α = 1 + α

Partial pressures of PCl_{5}, PCl_{3} and Cl_{2} will be:

p (PCl_{3}) = α p / 1 + α

p (Cl_{2}) = α p / 1 + α

p (PCl_{5}) = (1 – α) p / 1 + α

K_{p} = p (PCl_{3}) X p (Cl_{2}) / p (PCl_{5})

K_{p} = **[**(α p / 1 + α) X (α p / 1 + α)**]/[ **(1 – α) p / (1 + α)**]**

= α^{2} p / (1 – α) ^{2}

Substituting p = 1 atm and α = 0.2

K_{p} = (0.2) ^{2} X 1 / (1 – (0.2)) ^{2}

= 0.041

When α = 1/2 = 0.5, let pressure is p^{’}

K_{p} = α^{2} p^{’} / (1 – α) ^{2}

0.041 = (0.5) sup>2 p^{’} / (1 – (0.5)) ^{2}

p^{’} = (0.041) [1 – (0.5) ^{2}] / (0.5) ^{2}

p^{’} = 0.125 atm

# The degree of dissociation of PCl_{5} at a certain temperature and under atmospheric pressure is 0.2. Calculate the pressure at which it will be half dissociated at the same temperature.

**Category:** Free Energy and Chemical Equilibria

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