SbCl5 (g) SbCl3 (g) + Cl2 (g) Kp(450(degrees Celsius))= 1.5 If ...

SbCl5 (g) SbCl3 (g) + Cl2 (g) Kp(450(degrees Celsius))= 1.5 If you place 0.600 mol SbCl5 in a 2.0 L container at 450 degrees Celsius, determine the number of moles for each gas at equilibrium. T=450ºC= 723K. The relationship between the two equilibrium constants are:

Kp=Kc(RT)Δn where,

Δn = (Total moles of gas on the products side) - (Total moles of gas on the reactants side). Δn= 2-1=1 Then, in our problem

Now from the balanced reaction, we have 0.6 moles in 2L = 0.3mol/L SbCl5(g) ---->SbCl3(g)+Cl2(g) ni

0,3

0

0

nr

x

x

x

neq.

0.3-x

x

x

[

][ [

] ]

√

In the equilibrium. [SbCl5]= 0.3 – 0.075 = 0.225 mol/L [SbCl3]= x= 0.075mol / L [Cl2] = x = 0.075 mol/L In the 2L of mxture we have. Moles of SbCl5= 0.45 Moles of SbCl3= 0.15

Moles of Cl2= 0.15