How much PCl(5) must be added to a one l...

How much `PCl_(5)` must be added to a one little vessel at `250^(@)C` in order to obtain a concentration of 0.1 mole of `Cl_(2)` at equilibrium. `K_(c)` for `PCl(g)hArrPCl_(3)(g)+Cl_(2)(g)` is `0.0414M`

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How much PCl_(5) must be added to a one litre vesel at 250^(@)C in order to obtain concentration of 0.1 mole of Cl_(2) at equilibrium K_(C) for PCl_(5)(g) hArr PCl_(3(g)+Cl_(2) is 0.0414M

Initially 0.8 mole of PCl_(5) and 0.2 mole of PCl_(3) are mixed in one litre vessel . At equilibrium 0.4 mole of PCl_(3) is present . The value of K_(C) for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) is

PCl_(5) , PCl_(3) and Cl_(2) are at equilibrium at 500K and having concentration 1.59M PCl_(3) , 1.59M Cl_(2) and 1.41 M PCl_(5) . Calcualte K_(c) for the reaction PCl_(5)hArrPCl_(3)+Cl_(2)

PCl_(5),PCl_(3) and Cl_(2) are at equilibrium at 500K and having concentration 1.59M PCl_(3), 1.59M CL_(2) and 1.41 MPCl_(5) . Calculate K_(c) for the reaction PCl_(5)hArrPCl_(3)+Cl_(2)

One mole of N_(2) and 3 mole of PCI_(5) are placed in a 100 litre vessel heated to 227^(@) c. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, If K_(p) of the reaction PCl_(5) hArr PCl_(3)+Cl_(2) is y xx 10^(-1) then what is 'y'?

1 mol PCl_(3) and 1 mol Cl_2 are taken in a 10 lit vessel and heated . Now , if 0.8 mol PCl_(5) is present at equilibrium , then

What is the equilibrium concentration of each of the substances in the equilirbrium when the initial concentration of I Cl was 0.78M ? 2ICl(g)harrI_(2)(g)+Cl_(2)(g) , K_(c)=0.14

1 mol A_((g)) is heated in 1 lit closed vessel and equilibrium is reached at 300^(@) C in A_((g)) hArr B_((g)) . If K_(C) =4 , concentration of B_((g)) at equilibrium is (in mol/lit)

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