# if some he gas is introduced into equilibrium pcl5=pcl3+cl2 at const pressure and temp then kc will increase  decrease unchanged  none

Dear Student. Kc will be unchanged because it depends on temperature only. I am giving you some more detail. Consider the dissociation equilibrium of PCl5 : PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) According to law of chemical equilibrium , equilibrium constant (Kc) : Kc = [PCl3] [Cl2][PCl5] a) If an inert gas is added at constant volume then there is no effect on the dissociation of a compound as only the total pressure of the system increases and there is no change in the partial pressures of the gases. Hence , there will be not effect on the state of equilibrium. b) However, if an inert gas is added at constant pressure and the total volume (number of moles) increases, then in order to maintain total pressure and Kp same, the partial pressures have to adjust (i.e. decrease as volume and pressure are inversely proportional). Hence , the molar concentration of both reactants and products will decrease at equilibrium. There are two terms of concentration in the numerator and only one term of concentration in the denominator , therefore Kc should decrease. Kc is equilibrium constant at constant temperature. Hence, if we want to keep the Kc constant , either we have to decrease the [PCl5] or [PCl3] and we have to increase [Cl2] . This can only be achieved when more of PCl5 dissociates to give PCl3 and Cl2 . Hence, the dissociation increases with the addition of an inert gas. Number of moles increase on dissociation.​​ According to Le Chatelier's principle, a decrease in pressure due to increasing volume causes the reaction to shift to the side with more moles of gas. The equilibrium will move in such a way that the pressure increases again. It can do that by producing more molecules. Thus , the equilibrium would shift to the side with more gas molecules which is the right side and hence dissociation will increase. Regards

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