Question

the equilibrium composition for the reaction is pcl3+cl2=pcl5 pcl3=0.20 cl2= 0.10 pcl4=0.40 mole/litre what will be the equilibrium concentration of pcl5 after adding 0.10 mole of cl2 at same temperature

Answer 1

a) 0.61Kp = Kc× (RT)^∆n∆n=(no of moles of gaseous products - no of moles of gaseous reactants)Therefore, ∆n = 1-2= (-1)Now, Kc=26, R=0.0821, T= 250+273.15=523.15KHence, Kp= 26×(RT)^ -1=26/RT = 26/(0.0821×523.15) = 0.61

Answer 2

The equilibrium composition for the reaction is pcl3+cl2=pcl5 pcl3=0.20 cl2= 0.10 pcl4=0.40 mole/litre what will be the equilibrium concentration of pcl5 after adding 0.10 mole of cl2 at same temperature

Explanation:

We are given that :

PCl₃ +Cl₂--->PCl₅

0.20  0.10    0.40 mole/L

Kc= 0.40/0.20 x 0.10=20

New initial concentration of Cl₂=0.10 + 0.10 =0.20moles/litre

New initial concentration of PCl₅ and PCl₃ do not change .

Let us say that x moles of phosphorous penta chloride reacts , then :

Concentration of PCl₃=0.20 - x

Cl₂=0.20-x

PCl₅=0.40 + x

Now , Kc=[PCL₅]/[PCl₃][Cl₂]

that is :0.40+x /(0.20-x)(0.20-x)=20

or , x=0.05

hence [PCl₅]=0.40 + 0.05 = 0.45m/l