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Derivation of Kp and Kc

4.0 moles of ...

Question

$4.0$ moles of argon and $5.0$ moles of $P C l_{5}$ are introduced into an evacuated flask of $100$ litre capacity at $610 K$ . The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be $6.0 atm$ . The $K_{p}$ for the reaction is
[Given : $R = 0.082 L a t m K^{- 1} m o l^{- 1}$ ]

A

6.24

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B

12.13

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C

15.24

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D

2.25

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Solution

The correct option is D $2.25$
$P C l_{5} ⇋ P C l_{3} + C l_{2}$

$t = 0500$

$t = t 5 - n n n$

Total moles = 5 – n + n + n = 5 + n

For Argon
$n_{A r} = 4$
Total moles $= n_{A r} + n_{P C l_{5}} + n_{P C l_{3}} + n_{C l_{2}}$
$= 4 + 5 + n$
$= 9 + n$

$P V = n R T$
$6 \times 100 = (9 + n) \times 0.82 \times 610$
$n = 3$

$K_{p} = \frac{P_{P C l_{3}} . P_{C l_{2}}}{P_{P C l_{5}}}$

$\frac{(\frac{3}{12} \times 6) \times (\frac{3}{12} \times 6)}{\frac{2}{12} \times 6}$

$= \frac{27}{12} = \frac{9}{4} = 2.25 atm$

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500 K

0.10

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G i v e n : R = 0.082 {L atm K}^{- 1} m o l^{- 1}

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