Some applications of Gibbs energy: further understanding

Question 1

Gibbs energy of solution is:

Answer 1

ΔsolG⦵

used to predict the extent to which a salt is soluble

Question 2

If ΔsolG⦵ is negative, then

Answer 2

then the products are favoured at equilibrium and the salt is soluble

Question 3

if ΔsolG⦵ is positive, then

Answer 3

then the solid salt is favoured at equilibrium and so the salt is insoluble (or more accurately, sparingly soluble, as we have already identified that a process with a positive Gibbs energy change can take place to a certain extent as long as the magnitude of ΔG is not too large)

Question 4

the equilibrium constant for the following process is:
(aq)
M+X-(s) → M+(aq) + X-(aq)

Answer 4

K [M+(aq)][X-(aq)]

Question 5

the equilibrium constant can is called the ……………… and is given the symbol:

Answer 5

solubility product, Ksp

Question 6

the relationship between Ksp and ΔsolG⦵ is given by the expression:

Answer 6

ΔsolG⦵ = -RTlnKsp

Question 7

the relationship between ΔsolG⦵ , ΔsolH⦵ and ΔSsystem is:

Answer 7

ΔsolG⦵ = ΔsolH⦵ - TΔSsystem

Question 8

the larger the value of Ksp, the more ………………….. the salt

Answer 8

the more soluble the salt at 298K

Question 9

The following data for the solubility of Ca(NO3)2, ΔsolH⦵, ΔSsystem, TΔSsystem ΔsolG⦵ and Ksp is:

ΔsolH⦵ (J mol-1)             : -19 000
ΔSsystem (J K-1 mol-1) :  +45
TΔSsystem (J mol-1)     : +13 000
ΔsolG⦵ (J mol-1)          :  -32 000
Ksp                                :  410 000

use the data above the explain the solubility of calcium nitrate

Answer 9

ΔsolH⦵ is negative at -19 000 J mol-1 so the ΔhydrationH of the ions must be greater than ΔlatticeH of Ca(NO3)2

(Remember: ΔsolH[XY(s)] + ΔlatticeH[XY(s)] = ΔhyH[X+(g)] + ΔhydH[Y-(g)]
ΔSsystem is positive at +45 j K-1 Mol-1 so the increase in entropy produced by the breaking up of the lattice must be greater than the decrease in entropy of the water produced by the hydration of the ions (bond making is an exothermic process!)

since ΔsolH⦵ < 0 and ΔSsystem > 0 , ΔsolG⦵ is negative, so calcium nitrate is soluble in water as both enthalpy and entropy terms are favourable

ΔsolG⦵ incorporates both enthalpy and entropy!

Question 10

calculate ΔG at 298K for the dissociation of HCl acid in water:
ΔH = -58 kJ mol-1
ΔS = -55 J K-1 mol-1
for the following reaction:
HCl(aq) → H+(aq) + Cl-(aq)

Answer 10

ΔG = ΔH - TΔS
ΔG = -58000 - (298 x -55)
      =  -58000 - - 16390
      =  -41610 J mol-1
      = -41.6 kJ mol-1
ΔG is negative as the favourable enthalpy outweighs the unfavourable entropy

Question 11

rearrange ΔsolG⦵ = -RTlnKsp for Ksp

Answer 11

Ksp = e (-ΔG / RT)

where (-ΔG / RT) is a power of e

Question 12

rearrange ΔsolG⦵ = -RTlnKsp for Ksp, then use the following data to calculate K for the dissociation of HCl
ΔG = -41.6 kJ mol-1
T = 298 K
R = 8.31

Answer 12

Ksp = e (-ΔG / RT)
Ksp = e (- (-41.6 x 1000) / 8.31 x 298)
Ksp= 19,800,000

Question 13

calculate ΔG at 298K for the dissociation of HF acid in water:
ΔH = -9 kJ mol-1
ΔS = -71 J K-1 mol-1
for the following reaction:
HF(aq) → H+(aq) + F-(aq)

Answer 13

ΔG = ΔH - TΔS
ΔG = -9000 - (298 x -71)
ΔG = -9000 + 21158
ΔG = + 12158 J mol-1
ΔG = + 12.158 kJ mol-1

Question 14

Use the following data to find K for the dissociation of HF in water at 298K
ΔG = + 12.158 kJ mol-1

Answer 14

K = e (-ΔG / RT)
K = e ( - (12.158 x 1000) / 8.31 x 298)
K = 0.00738