Question

Part A

Given the two reactions

H2S⇌HS−+H+,   K1 = 9.11×10⁻⁸, and

HS−⇌S2−+H+,   K2 = 1.01×10⁻¹⁹,

what is the equilibrium constant Kfinal for the following reaction?

S2−+2H+⇌H2S

Enter your answer numerically.

Part B

Given the two reactions

PbCl2⇌Pb2++2Cl−,   K3 = 1.89×10⁻¹⁰, and

AgCl⇌Ag++Cl−,   K4 = 1.26×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+⇌2AgCl+Pb2+

Express your answer numerically.

Part C

Part B

Given the two reactions

PbCl2⇌Pb2++2Cl−,   K3 = 1.89×10⁻¹⁰, and

AgCl⇌Ag++Cl−,   K4 = 1.26×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+⇌2AgCl+Pb2+

Express your answer numerically.

Part D

Part B

For the reaction

X(g)+3Y(g)⇌2Z(g)

Kp = 1.78×10⁻² at a temperature of 127 ∘C .

Calculate the value of Kc.

Express your answer numerically.

Answer 1

1.Given reactions are

H2S<--> HS- + H+, K1= 9.11*10^-8 , K1= [HS-][H+]/[H2S] -(1)

HS-< ---> S2-+H+, K2= 1.01*10^-19,   K2= [S2-] [H+]/[HS-] (2)

Eq.2*Eq.1 gives K1K2=K=   [H+]²[S2-]/[H2S]= 9.11*1.01*10^-27 = 9.2*10^-27

for the given reaction, S2-+2H+ <->H2S

K’= [H2S]/ [S2-] [H+]²= 1/K= 10²⁷/ 9.2=1.086*10²⁶

2^.

Given the two reactions

PbCl2⇌Pb2++2Cl−,   K3= [Pb+2] [Cl-]²/ [PbCl2] = 1.89×10⁻¹⁰    (1)

and

AgCl⇌Ag++Cl−,   K4 = = [Ag+] [Cl-]/[AgCl]= 1.26×10⁻⁴                 (2)

Squaring K4,

K4²= [Ag+]² [Cl-]²/ [AgCl]²= (1.26*10^-4)²=1.5876*10^-8 –(2A)

Eq.2A/ Eq,1 =

Kf= [Ag+]²[PbCl2]/ [Pb+2] [AgCl]² =1.5876*10^-10/(1.89*10^-10)=84

But for the given reaction, PbCl2+2Ag+⇌2AgCl+Pb2+

K’= [AgCl]2[ Pb+2] /[Ag+]2[ [PbCl]= 1/Kf= 1/84= 0.0119

3. X(g)+3Y(g)<----->2Z(g)

Kp= Kc*(RT)^deltan, deltan= change in moles during the reaction = 2-1-3=-2

R=0.0821 L.atm/mole.K, T= 127 deg.c= 127+273= 400K

KP=1.78*10-2= Kc*(0.0821*400)^-2,

KC= 1.78*10^-2/0.000927= 19.19