`PCl_(5)(g)rarr PCl_(3)(g)+Cl_(2)(g)` for this, equilibrium constant `K_(c)=1.8times10^(-7)` at `298K`. Calculate `Delta G^(0)` for the reaction.

For the equilibrium 2NOCl(g) hArr 2NO(g)+Cl_(2)(g) the value of the equilibrium constant, K_(c) is 3.75 xx 10^(-6) at 1069 K . Calcualate the K_(p) for the reaction at this temperature?

For the chemical equilibrium PCl_(5)(g) hArr PCl_(3)(g) +Cl_(2) (g) " at " 298K, K_(c )=1.8xx10^(-7) . Calculate Delta_(r )G^(Theta) for the forward reaction R=8.31JK^(-1)"mol"^(-1)

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.

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

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

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

Phosphorous pentachloride when heated in a sealed tube at 700 K it undergoes decomposition as PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g), K_(p)=38 atm Vapour density of the mixture is 74.25 . Equilibrium constant K_(c) for the reaction will be

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

For the reaction, 2NOCl(g) hArr 2NO(g)+Cl_(2)(g) Calculate the standard equilibrium constant at 298 K . Given that the value of DeltaH^(ɵ) and DeltaS^(ɵ) of the reaction at 298 K are 77.2 kJ mol^(-1) and 122 J K^(-1) mol^(-1) .

PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , In above reaction, at equilibrium condition mole fraction of PCl_(5) is 0.4 and mole fraction of Cl_(2) is 0.3. Then find out mole fraction of PCl_(3)