In: Chemistry
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.
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