The equilibrium pressure necessary to obtain `50%` dissociation of `PCl_5` at `250^@C` is numerically________of `K_p`.

Prove that the pressure necessary to obtain 50% dissociation of PCl_(5) at 250^(@)C is numerically three times of K_(p) .

Prove that the pressure necessary to obtain 50% dissociation of PCl_(5) at 250^(@)C is numerically three times of K_(p) .

At 300^@C the pressure necessary to obtained 80% dissociation of PCl_5 is numerically equal to

At 300^@C the pressure necessary to obtained 80% dissociation of PCl_5 is numerically equal to

Prove that the pressure necessary to obtain 50% dissociation of PCl_(5) at 500 K is numerically three times the value of K_(p) .

Total equilibrium pressure necessary to obtain 60% dissociation of PCl_5 is P at 500 K. The relation between equilibrium constant K_p and P is

For PCl_5(g)⇌PCl_3(g)+Cl_2(g) at equilibrium, K_p = P/3 , where P is equilibrium pressure. Then degree of dissociation of PCl_5 at that temperature is ?

The equilibrium constant K_(p) for the thermal dissociation of PCl_(5) at 200^(@)C is 1.6 atm. The pressure (in atm) at which it is 50% dissociated at that temperature is