JEE Advanced 2020 Paper 2 · Chemistry · Chemical Equilibrium · easy

For the gaseous equilibrium PCl5(g)rightleftharpoons PCl3(g)+Cl2(g) the value of Kp at a certain temperature is 1.0 atm. Pure PCl5 is allowed to…

Numerical answer

For the gaseous equilibrium

\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}

the value of

K_p

at a certain temperature is

1.0\ \mathrm{atm}

. Pure

\mathrm{PCl_5}

is allowed to dissociate at this temperature. The total equilibrium pressure, in atm, at which

\mathrm{PCl_5}

is exactly

50\%

dissociated is ______.

Show answer and solution

Answer: 3

Start with

1

mol of

\mathrm{PCl_5}

and degree of dissociation

\alpha

. At equilibrium the amounts are

(1-\alpha)

\mathrm{PCl_5}

and

\alpha

each of

\mathrm{PCl_3}

and

\mathrm{Cl_2}

, giving

(1+\alpha)

mol in total. For a total pressure

P

the partial pressures are

\dfrac{(1-\alpha)P}{1+\alpha}

for

\mathrm{PCl_5}

and

\dfrac{\alpha P}{1+\alpha}

for each product. Hence

K_p=\dfrac{\alpha^2P}{(1+\alpha)(1-\alpha)}=\dfrac{\alpha^2P}{1-\alpha^2}

. Putting

\alpha=0.5

gives

\dfrac{0.25P}{0.75}=\dfrac{P}{3}

. Setting

K_p=1.0\ \mathrm{atm}

gives

P=3\ \mathrm{atm}

. Check:

\mathrm{PCl_5}

is at

1\ \mathrm{atm}

and each product at

1\ \mathrm{atm}

, so

K_p=\dfrac{1\times1}{1}=1

. A common mistake is to use

K_p=\alpha^2P/(1-\alpha)

, which would give

P=1

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