Question

If Kp=3.13x10^-2 at 500 K for the reaction:

Br₂ (g) + Cl₂ (g) <--> 2 BrCl (g)

a) Calculate K_c at 500K
b) What is the value of Kc for the reverse reaction at 500 K?
c) Calculate the value of Kp for the following reaction at 500 K:
   2Br₂ (g) + 2Cl₂ (g) <--> 4BrCl (g)
d) Given the following two reactions and their equilibrium constants:
   AgCl (s)<--> Ag⁺¹ (aq) + Cl^-1 (aq) Kc=1.8x10^-10 M²
   Ag⁺¹ (aq) + 2NH₃ (aq) <--> [Ag(NH₃)₂]⁺¹   Kc=1.7x10⁺⁷ M-2
What is the equilibrium expression and value of Kc for the following reaction?
   AgCl (s) + 2NH₃ (aq) <--> [Ag(NH₃)₂]⁺¹ (aq) + Cl^-1 (aq)

Answer 1

a)Br₂ (g) + Cl₂ (g) <--> 2 BrCl (g)

Kp=Kc(RT)ⁿ

n=(mole of gaseous product)-(mole of gaseous reactant)=(2)-(1+1)=0

so K_p=K_c=3.13*10^-2

b)2 BrCl (g)<------->Br₂ (g) + Cl₂ (g)

for reverse of a reaction

c)2Br₂ (g) + 2Cl₂ (g) <--> 4BrCl (g)

reaction is multipiled by n=2

d)AgCl (s)<--> Ag⁺¹ (aq) + Cl^-1 (aq) Kc=1.8x10^-10 M²
   Ag⁺¹ (aq) + 2NH₃ (aq) <--> [Ag(NH₃)₂]⁺¹   Kc=1.7x10⁺⁷ M-2

the rection obtained by adding above two reaction is

AgCl (s) + 2NH₃ (aq) <--> [Ag(NH₃)₂]⁺¹ (aq) + Cl^-1 (aq)

so K_c^'=multiplication of both equlibrium constant=1.8*10^-10*1.7*10⁷=3.06*10^-3