At `250^@C`, `K_c` for `PCl_5(g)hArr PCl_3(g)+Cl_2(g)` is `0.04`. How many moles of `PCl_5` must be added to a `3-L` flask to obtain `0.15M Cl_2` at equilibrium?

At 25^(@)C , the thermal decomposition of PCl_5 occuring in a closed vessel of volume 1L gives rise to the equilibrium: PCl_5(g)hArr PCl_3(g)+Cl_2(g) . How many moles of PCl_5 is to be added so that the concentration of Cl_2(g) at equilibrium will be 0.1mol*L^(-1) ? [Given : K_p=1.78 (at 250^(@)C ]

Consider a reaction PCl_5(g) rarr PCl_3(g)+Cl_2(g) b) Explain why does PCl_5 decomposes easily?

A quantity of PCl_5 was heated in a 10 litre vessel at 250^@ C to show PCL_5(g) hArr PCl_3 + Cl_2 AT equilibrium the vessel contains 0.1 mole of PCl_5 0.20 mole of PCl_3 and 0.20 mole od cl_2 The equilibrium constant of the reaction is :

A quantity of PCl_(5) was heated in a 10 dm^(3) vessel at 250^(@)C PCl_(5(g)) hArr PCl_(3 (g)) + Cl_(2(g)) . At equilibrium the vessel contains 0.1 mole of PCl_(5) , 0.2 moles of PCl_(3) and 0.2 moles of Cl_2 . The equilibrium constant of the reaction is