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Stoichiometry

K C for P Cl ...

Question

$K_{c}$ for $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ is 0.45 at $250^{\circ} C$ . How many mole of $P C l_{5}$ must be added to a 3 - litre flask to obtain a $C l_{2}$ concentration of 0.15 M?

0.45

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0.6

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1.35

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1.8

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Solution

The correct option is B 0.6
Here $K_{c} = 0.45. [C l_{2}] = 0.15 M$
No. of moles of $C l_{2}$ at equrilibrium concentration is $0.15 \times 3 = 0.45 moles .$
Let the no of moles of $P C l_{5}$ added is 'x'
$P C l_{5} ⇌ P C l_{3} + C l_{2}$

Then,
$(\frac{x - 0.45}{3})$ will be the equilibrium concentration of $P C l_{5}$
$∴ K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = 0.45$
$\Rightarrow \frac{0.15 \times 0.15}{(\frac{x - 0.45}{3})} = 0.45$
$\Rightarrow x = 0.6 moles$

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Kc for PCl5(g)⇌PCl3(g)+Cl2(g) is 0.45 at 250∘C. How many mole of PCl5 must be added to a 3 - litre flask to obtain a Cl2 concentration of 0.15 M?

$K_{c}$ for $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ is 0.45 at $250^{\circ} C$ . How many mole of $P C l_{5}$ must be added to a 3 - litre flask to obtain a $C l_{2}$ concentration of 0.15 M?