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Byju's Answer

Standard XII

Chemistry

Reaction Quotient

In the reacti...

Question

In the reaction $P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ the equilibrium concentration of $P C l_{5}$ and $P C l_{3}$ are 0.4 and 0.2 mole/litre respectively. If the value of $K_{C}$ is 0.5 what is concentration of $C l_{2}$ in moles/litre ?

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Solution

The correct option is C 1
The equilibrium reaction is $P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ .
The value of the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$ .
But, $[P C l_{5}] = 0.4 M, [P C l_{3}] = 0.2 M, K_{c} = 0.5$ .
Substitute values in the above expression.
$0.5 = \frac{0.2 \times [C l_{2}]}{0.4]}$
Thus, $[C l_{2}] = 1.0$ .

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Similar questions

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In the reaction PCl5(g)⇔PCl3(g)+Cl2(g) the equilibrium concentration of PCl5 and PCl3 are 0.4 and 0.2 mole/litre respectively. If the value of KC is 0.5 what is concentration of Cl2 in moles/litre ?

The correct option is C 1The equilibrium reaction is PCl5(g)⇔PCl3(g)+Cl2(g).The value of the equilibrium constant is Kc=[PCl3][Cl2][PCl5].But, [PCl5]=0.4M,[PCl3]=0.2M,Kc=0.5.Substitute values in the above expression.0.5=0.2×[Cl2]0.4]Thus, [Cl2]=1.0.

In the reaction $P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ the equilibrium concentration of $P C l_{5}$ and $P C l_{3}$ are 0.4 and 0.2 mole/litre respectively. If the value of $K_{C}$ is 0.5 what is concentration of $C l_{2}$ in moles/litre ?