The reaction, `PC1_(5)hArrPC1_(3) + C1_(2)` is started in a 5 litre container by taking one mole of `PC1_(5)`. If `0.3` mole of `PC1_(5)` is there at equilibrium, concentration of `PC1_(3)` and `K_(c)` will respectively be:

The reaction, PCI_(5)hArrPCI_(3)+CI_(2) is started in a five litre container by taking one mole of PCI_(5) . If 0.3 mol PCI_(5) is there at equilibrium, concentration of PCI_(3) "and" K_(c) will respectively be:

PCl_5 to PCl_3 + Cl_2 K_c = 1.844 If 3 moles of PCl_5 are taken in 1L vessel. Find equilibrium concentration of PCl_5

In 5 litre container 1 mole of H_(2) and 1 mole of I_(2) are taken initially and at equilibrium of HI is 40% . Then K_(p) will be

In 5 litre container 1 mole of H_(2) and 1 mole of I_(2) are taken initially and at equilibrium of HI is 40% . Then K_(p) will be

If dissociation for reaction, PC1_(5)hArrPC1_(3) + C1_(2) is 20% at 1 atm pressure. Calculate K_(c) .

Suppose the reaction PC1_(5(s))hArrPC1_(3(s)) + C1_(2(g)) is in a closed vessel at equilibrium stage. What is the effect on equilibrium concentration of C1_(2(g)) by adding PC1_(5) at constant temperqture ?

Suppose the reaction PC1_(5(s))hArrPC1_(3(s)) + C1_(2(g)) is in a closed vessel at equilibrium stage. What is the effect on equilibrium concentration of C1_(2(g)) by adding PCl_(5) at constant temperature ?

A quantity of PCI_(5) was heated in a 10 litre vessel at 250^(@)C, PCI_(5)(g)hArrPCI_(3)(g) + CI_(2)(g) . At equilibrium the vessel contains 0.1 mole of PCI_(5), 0.20 mole of PCI_(3) and 0.2 mole of CI_(2) . The equilibrium constant of the reaction is