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Standard XII

Chemistry

Derivation of Kp and Kc

Pure PCl5 is ...

Question

Pure $P C l_{5}$ is introduced into an evacuated chamber and comes to equilibrium at $247^{\circ} C$ and $2.0 a t m$ . The equilibrium gaseous mixture contains $40 %$ chlorine by volume. Calculate $K_{p}$ at $247^{\circ} C$ for the reaction
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$

0.625 a t m

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4 a t m

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1.6 a t m

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None of these

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Solution

The correct option is C $1.6 a t m$
The gaseous mixture contains $40 % C l_{2}$ and $40 % P C l_{3}$ , since they are produced in $1 : 1$ mole ratio. The $P C l_{5}$ is $20$ .
For ideal gases mole % = volume %
$P_{C l_{2}} = P_{P C l_{3}}$
$\Rightarrow 2 \times 0.40 = 0.80 atm$
$P C l_{5} = 2 \times 0.2 = 0.40 atm$
$∴ K_{p} = \frac{P_{P C l_{3}} . P_{C l_{2}}}{P_{P C l_{5}}}$
$= \frac{0.80 \times 0.80}{0.40}$
$= 1.6 atm$

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Pure PCl5 is introduced into an evacuated chamber and comes to equilibrium at 247∘C and 2.0 atm. The equilibrium gaseous mixture contains 40% chlorine by volume. Calculate Kp at 247∘C for the reaction PCl5(g)⇌PCl3(g)+Cl2(g)

The correct option is C 1.6 atmThe gaseous mixture contains 40% Cl2 and 40 %PCl3, since they are produced in 1:1 mole ratio. The PCl5 is 20. For ideal gases mole % = volume % PCl2=PPCl3 ⇒2×0.40=0.80 atm PCl5=2×0.2=0.40 atm ∴Kp=PPCl3.PCl2PPCl5 =0.80×0.800.40 =1.6 atm

Pure $P C l_{5}$ is introduced into an evacuated chamber and comes to equilibrium at $247^{\circ} C$ and $2.0 a t m$ . The equilibrium gaseous mixture contains $40 %$ chlorine by volume. Calculate $K_{p}$ at $247^{\circ} C$ for the reaction
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$