Thermodynamics II Flashcards
Gibbs free energy
G = H – TS therefore ∆G = ∆H - T ∆S
What does each letter mean?
G = Gibbs Free Energy (kJ mol-1)
H = Enthalpy (kJ mol-1)
S = Entropy (J K-1 mol-1)
T = Temperature (K)
Gibbs Free Energy can inform us if the reaction is_________ or not (Will it occur of it’s own accord)
spontaneous
∆G negative:
Spontaneous in Forward direction only (irreversible)
∆G zero:
Reaction at equilibrium, can proceed in either direction (reversible)
∆G positive:
Non-spontaneous, will not proceed in the forward direction unless coupled with an energetically favourable reaction.
Coupled reactions:
Other reaction must have a larger, negative ∆G value. Sometimes called tandem reactions.
T is what units?
Kelvin
C > K = +273(.15)
Endothermic +∆H / -∆S, spon?
+∆G always
Spontaneous = Never
Endothermic +∆H / +∆S, spon?
∆G + low T, - high T
Spontaneous = on heating
Exothemic -∆H / +∆S, spon?
-∆G always
Spon = Always
Exothermic -∆H / -∆S, spon?
+∆G (high T), - (low T)
Spon = on cooling
Free Energy of Formation (∆Gf⁰)
The free energy change which results from 1 mol of substance prepared from its elements at standard pressure (1 atm) and a given temperature (usually 298 K). The units are kJ/mol
The ∆Gf⁰ of chemical elements is always ____
0 = zero
What does the little 0 with a line through mean?
In standard conditions
Oxidative Metabolism (cellular respiration)
Process by which nutrients from our diet are converted into energy
There are several intermediate steps
- Glycolysis + Kreb’s cycle - but overall reaction for cellular oxidation of glucose is:
∆G = ∆H - T ∆S
Glucose + 6O2 —> 6CO2 + 6H2O
Is this process endo/ excothermic?
Exothermic
Glucose + 6O2 —> 6CO2 + 6H2O
Increase or decrease in entropy?
Increase
Goes from 7 molecules (gas and solid) to 12 (liquid and gas molecules)
Glucose + 6O2 —> 6CO2 + 6H2O
Spontaneous reaction?
∆G = ((6 x -394) + (6 x 2370) – (-917.2) + 0)
∆G = -2871.4 kJ/mol
-∆H +∆S
∆G will always be -ive
