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0.6 mole of PCl(5), 0.3 mole of PCl(3) a...

0.6 mole of `PCl_(5)`, 0.3 mole of `PCl_(3)` and 0.5 mole of `Cl_(2)` are taken in a 1 L flask to obtain the following equilibrium :
`PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`
If the equilibrium constant `K_(c)` for the reaction is 0.2 Predict the direction of the reaction.

A

Forward direction

B

Backward direction

C

Direction of the reaction cannot be predicted

D

Reaction does not move in any direction

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The correct Answer is:
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0.6 moles of PCl_(5) , 0.3 mole of PCl_(3) and 0.5 mole of Cl_(2) are taken in a 1 L flask to obtain the following equilibrium , PCl_(5(g))rArrPCl_(3(g))+Cl_(2(g)) If the equilibrium constant K_(c) for the reaction is 0.2 Predict the direction of the reaction.

(K_p/K_c) for the given equilibrium is [PCl_5 (g) hArrPCl_3(g) + Cl_2 (g)]

For the equilibrium PCl_5(g)hArr PCl_3(g) +Cl_2 , the value of the equilibrium constant decreases on additions of Cl_2 gas.

Consider the following eqilibrium PCl5(g))--->PCl_((3)(g)) + Cl_((2)(g)), What happens to the equillibrium when concentration of Cl2 is increased

Find out the units of K_c and K_p for the following equilibrium reactions : PCl_5(g) hArr PCl_3 + Cl_2(g)

For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

For PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g), write the expression of K_(c)

n mole of PCl_(3) and n mole of Cl_(2) are allowed to react at constant temperature T to have a total equilibrium pressure P , as : PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g) If y mole of PCl_(5) are formed at equilibrium , find K_(p) for the given reaction .

For the reaction PCl_(5)(g) rightarrow PCl_(3)(g) + Cl_(2)(g)

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