Cards (62)

  • System
    Our interest
  • Surroundings
    The rest of the universe
  • Gibbs free energy
    Tells us if a change to our system will happen spontaneously or not
  • dG
    Change in free energy
  • ΔH
    Change in enthalpy
  • ΔS
    Change in entropy
  • T
    Temperature in Kelvin
  • Mass action ratio
    q = [C][D]/[A][B]
  • Mass action ratio
    • F-1,6-bP ⇌ G3P + DHP
  • ΔG⁰
    Standard Free Energy
  • R
    Gas constant
  • T
    Absolute temperature
  • q
    Mass action ratio
  • ΔG⁰ is the change in Free Energy when concentration of all reactants and products are 1M at 25°C, 1 atm.
  • ΔG⁰' is the Standard Free Energy at pH 7 at 25°C and 1 atm
  • ΔG depends on the nature of the reaction and the concentrations
  • If ΔG = 0
    No free energy change takes place, the system is at equilibrium
  • Rate up

    Rate down
  • Number of cars in
    Number of cars out
  • Rate of degradation

    Rate of formation of F-1,6-bP
  • The equilibrium constant is related to the standard free energy in a logarithmic manner
  • R=8.31x10^-3 kJ mol^-1 K^-1
  • T=298 K
  • Relationship between ΔG⁰' and Keq

    • ΔG⁰' = -5.7 kJ/mol, Keq = 10
    • ΔG⁰' = -11.5 kJ/mol, Keq = 100
    • ΔG⁰' = -17.3 kJ/mol, Keq = 1000
  • Relatively small changes in ΔG⁰' can lead to big changes in Keq
  • Conversion of dihydroxyacetone-phosphate to glyceraldehyde-3-phosphate

    [G3P]/[DHAP] = 0.0475 at equilibrium at 25°C
  • In vivo [DHAP] = 2 x 10^-4 M and [G3P] = 3 x 10^-6 M
  • ATP is the main energy carrier in the cell
  • Pi
    Inorganic phosphate
  • Adenylate Kinase interconverts ATP, ADP and AMP
    ATP + AMP ⇌ 2 ADP
  • In the body more energy is released as the reaction is not at equilibrium
  • A cheetah running 110 km/h uses 55g ATP/s
  • A cheetah weighs 20-70 kg
  • A 10 s sprint uses 0.55 kg ATP or 3% of body weight for a small cheetah
  • Humans apparently need about 70 kg ATP/day
  • Glucose + Pi ⇌ Glucose-6-phosphate + H2O

    ΔG⁰' = +14.0 kJ/mol, Keq = 3.5 x 10^-3
  • ATP + H2O ⇌ ADP + Pi

    ΔG⁰' = -30.5 kJ/mol
  • Glucose + ATP ⇌ Glucose-6-phosphate + ADP

    ΔG⁰' = -16.5 kJ/mol, Keq = 8.0 x 10^2
  • Coupling to ATP shifts the equilibrium 5 orders of magnitude
  • This only works because Pi is a shared intermediate