The Second Law and Gibbs energy

Question 1

the second law of thermodynamics states that

Answer 1

the total entropy of an isolated system cannot decrease over time.

Question 2

Gibbs energy is a

Answer 2

quantity used for convenience, related to the entropy change of the system

Question 3

Gibbs energy=

Answer 3

ΔG = ΔH - TΔSsystem
where T is temperature of the system in Kelvin
note this is the change in the entropy of teh System not the change in temperature (under standard conditions T will always be 298)
ΔH is measured in kJ while ΔSsystems is in J, so must divide by 1000

Question 4

when ΔG is negative, a process or reaction is thermodynamically

Answer 4

feasable

Question 5

when ΔG is positive, a process or reaction is thermodynamically

Answer 5

unfeasable

Question 6

if ΔG is zero, a process or reaction is

Answer 6

in equilibrium

Question 7

ΔG will be negative when:

Answer 7

ΔH < 0 and ΔSsystem > 0
ΔH < 0, ΔSsystem < 0 but the magnitude of ΔH > the magnitude of TΔSsystem
ΔH > 0, ΔSsystem > 0 but the magnitude of ΔH < the magnitude of TΔSsystem

Question 8

ΔG will always be positive when:

Answer 8

ΔH > 0 and ΔSsystem < 0
ΔH > 0, ΔSsystem > 0 but the magnitude of ΔH > the magnitude of TΔSsystem
ΔH < 0, ΔSsystem < 0 but the magnitude of ΔH the magnitude of TΔSsystem

Question 9

calculate, using Gibbs energy, whether water can freeze at +5° C or at -5° C
H2O(l) → H2O(s)
ΔH= -6010 J mol-1
ΔSsystem = -22.0 J K-1 mol-1

Answer 9

ΔG = ΔH - TΔSsystem
At +5° C
ΔG= -6010 - ((273 +5) x -22.0) = +106 J mol-1
ΔG is positive, so reaction is not feasible at 5° C
At -5° C
ΔG = -6010 - ((273 -5) x -22.0) = -114 J mol-1
Here ΔG is negative, so water will freeze at -5° C

Question 10

the units for Gibbs energy are:

Answer 10

J mol-1

Question 11

when ΔG = 0, then

Answer 11

ΔH = TΔSsystem

Question 12

when ΔH = TΔSsystem, then ΔG=

Answer 12

ΔG = 0

Question 13

Determine the minimum temperature at which the thermal decomposition of calcium carbonate becomes feasible.
CaCO3(s) → CaO(s) + CO2(g)
ΔH = +178 kJ mol-1
the entropies of CaCO3(s) = 89, CaO(s) = 40 and CO2(g) = 214 J K-1 mol-1.

Answer 13

1) ΔG will have to equal 0 for the minimum temp. at which CaCO3 will decompose, so when ΔG=0, then
ΔH = TΔSsystem, where T= ΔH / ΔSsystem
2) ΔSsystem = Total entropy of products - Total entropy of reactants, so ΔSsystem = 40+214-(+89) = 165 J K-1 mol-1
3) T = ΔH / ΔSsystem = 178 x 1000 / 165 = 1079 K