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Derivation of Kp and Kc

2 moles of ...

Question

$2$ moles of $P C l_{5}$ were heated in a closed vessel of 2 litre capacity. At equilibrium, $40 %$ of $P C l_{5}$ is dissociated into $P C l_{3}$ and $C l_{2}$ . The value of equilibrium constant is-

0.266

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0.53

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2.66

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5.3

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Solution

The correct option is B $0.266$

$P C l_{5}$
$P C l_{3}$
$C l_{2}$
Initial
2
0
0
Change
-0.8
0.8
0.8
Equilibrium
1.2
0.8
0.8
Molar concentration
0.6
0.4
0.4

The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{0.4 \times 0.4}{0.6} = 0.266.$

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Similar questions

2 moles of $P C l_{5}$ were heated in a closed vessel of 2 litre capacity. At

equilibrium, 40% of $P C l_{5}$ is dissociated into $P C l_{3}$ and $C l_{2}$ . The value of

equilibrium constant is

P C l_{5},

P C l_{5}

P C l_{3} a n d C l_{2} .

P C l_{5}

P C l_{5}

P C l_{3}

C l_{2}

2

P C l_{5}

2

40 %

P C l_{5}

P C l_{3}

C l_{2}

P C l_{5}

P C l_{5}

P C l_{3}

C l_{2}

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