Energy
20.3, Free 20.4
Last time
G = H - TS G is negative means spontaneous G is positive means non-spontaneous S is negative means order S is positive means disorder
Example
Is this reaction spontaneous?
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) A. Always Spontaneous B. Depends on Temperature C. Never Spontaneous
If you put CO2 and H2O in a room long enough, will you ever get propane?
Example
Solve for the G of this reaction at T = 25 Celsius: S(g) + O2(g) SO2(g) H = -575.8 kJ/mol S = -124.9 J/mol-K
Another Example
Solve for the G of this reaction at T = 25C: H2O(l) H2O(s) H = -6.012 kJ/mol S = -22.0 J/mol-K At what T does G = 0?
Next point
When G = 0, the system is at equilibrium
Both the forward and reverse occur
When G < 0, spontaneous reaction, the forward reaction dominates When G > 0, non-spontaneous reaction, the reverse reaction dominates
More G
You can also find G by: G = G0(products) G0(reactants) Note: This only works at T = 25 Celsius
Example
Mg2+(aq) + 2 Cl-(aq) MgCl2(s) Solve for G at T = 25 Celsius
G0(Mg2+) = -456.1 G0(Cl-) = -131.7
G0(MgCl2) = -592.1
Work
Spontaneous reactions can do work, the most work a reaction can do equals G Non-spontaneous reactions require work in order to occur. Each non-spontaneous reaction requires at least G amount of work How much work can 8 g of S in the reaction: S(g) + O2(g) SO2(g) do at 25 Celsius?
Coupling
If a reaction is non-spontaneous will it occur?
ever By itself, without an outside force, no But the reaction can occur if the energy is supplied from another reaction
Cu2O(s) 2Cu(s) + O2(g) C(s) + O2(g) CO(g)
G = 140 G = -143.8
Cu2O(s) + C(s) 2Cu(s) + CO(g) G = -3.8
Result
Thus Cu2O(s) will not spontaneously turn into Cu and O2 But with C(s) present, it will turn into Cu and CO Hess's Law still works
Back to K
G
G is negative
Forward reaction
Occurs more than the reverse Equilibrium with Reverse reaction Occurs less than the reverse (nonspontaneous)
K
Q < K reaction proceeds to the right Q = K equilibrium reached Q > K reaction proceeds to the left
G = 0 G is positive
Back to K
G = - RT ln K
0
R = 8.314 T is in Kelvin G in J/mol
This equation (above) works for only the standard state (everything at 1M or at 1atm) Usually we use this to solve for K at any temperature
G = G + RT ln Q
0
We've always solved for the G0 (standard state) before But if we have different concentrations to begin with (e.g. a value for Q), then the G is different as well Usually we use this equation to solve for G when G0 and Q are known
Example
Reaction: Pb2+(aq) + 2 Cl-(aq) PbCl2(s) Hf (kJ/mol) = -26.6 S0 (J/mol-K) = 5
Solve for G0 and K at T = 298 Kelvin
Same Example
Solve for G if we begin with 0.1 M Pb2+ and 0.5 M Cl- at 298 Kelvin Compare Q at these conditions to K. Would you expect the reaction to be spontaneous at these conditions based on this comparison?