Question

Part A

Given the two reactions

H2S⇌HS−+H+,   K1 = 9.20×10⁻⁸, and

HS−⇌S2−+H+,   K2 = 1.18×10⁻¹⁹,

what is the equilibrium constant Kfinal for the following reaction?

S2−+2H+⇌H2S

Enter your answer numerically.

Kfinal =

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Part B

Given the two reactions

PbCl2⇌Pb2++2Cl−,   K3 = 1.75×10⁻¹⁰, and

AgCl⇌Ag++Cl−,   K4 = 1.28×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+⇌2AgCl+Pb2+

Express your answer numerically.

Answer 1

A)

we need S-2 and 2 H+ ions in the left so invert equation (1 and 2); when inverting, we must divide K2 --> 1/K2

HS−+H+⇌H2S K1 = 1/(9.20×10−8)

S2−+H+⇌HS− K2 = 1/(1.18×10−19)

Total K

K = (1/(9.20*10^−8))(1/(1.18*10^−19)) = 9.211*10^25 (strongly favoured)

B)

PbCl2⇌Pb2++2Cl−,   K3 = 1.75×10−10,

AgCl⇌Ag++Cl−,   K4 = 1.28×10−4,

we need

2 times Ag+ so... multiply (2) by 2 K4 ---> turns to --> K4^2

PbCl2⇌Pb2++2Cl−,   K3 = 1.75*10^−10,

2AgCl⇌2Ag++2Cl−,   K4 = (1.28*10^−4)^2

now, we need (2) inverted so... K4 --> 1/K4

PbCl2⇌Pb2++2Cl−,   K3 = 1.75*10^−10,

2Ag++2Cl−⇌2AgCl K4 = (1.28*10^−4)^-2

then

K = (1.75*10^−10)((1.28*10^−4)^-2) = 0.01068115234

slighlty favours reactants