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Standard XIII

Chemistry

Introduction to Le Chatelier's Principle

The equilibri...

Question

The equilibrium constant for the reaction $B r_{2} (l) + C l_{2} (g) ⇌ 2 B r C l (g)$ at $27^{\circ}$ C is $K_{p} = 1 atm$ . In a closed container of volume $164 L$ initially 10 moles of $C l_{2}$ are present at $27^{\circ} C$ . What minimum moles of $B r_{2} (l)$ must be introduced into this container so that the above equilibrium is maintained at a total pressure of 2.25 atm? Vapour pressure of $B r_{2} (l)$ at $27^{\circ}$ C is 0.25 atm. Assume that volume occupied by the liquid is negligible.
[ $R = 0.082 {Latmmol}^{- 1} K^{- 1}$ ,Atomic mass of bromine = 80]

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Solution

To maintain equilibrium there must be some moles of $B r_{2} (l)$ at equilibrium
$B r_{2} (l) ⇌ B r_{2} (g)$

$n = \frac{P V}{R T}$

$n = \frac{(0.25) (164)}{(0.082) (300)} = 1.667$

Now for equilibrium
$B r_{2} (l) + C l_{2} ⇌ 2 B r C l (g)$
x 10
10-x 2x

Total gaseous moles = 10+x
We have $P_{C l_{2}} + P_{B r C l} = 2 a t m$
so
$P_{C l_{2}}$ = $\frac{(10 - x)}{(10 + x)} \times 2 a t m$

$P_{B r C l}$ = $\frac{(2 x)}{(10 + x)} \times 2 a t m$

$K_{p} = \frac{p_{B r C l}^{2}}{p_{C l_{2}}}$
on putting values and solving
x = 3.333
Hence total moles of $B r_{2} (l)$ required to maintain both equilibria = 1.667 + 3.333 = 5

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