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CO_4
Biochemical Thermodynamics
Dr. Koteswara Reddy Gujjula
Overview of the Presentation
 Biochemical Stoichiometry
 Equilibrium Constant and Gibbs Free Energy
 Growth Stoichiometry and Elemental Balance
 Biomass Yield and Product Yield
 Laws of Thermodynamics in Biochemical
Reactions
Biochemical Stoichiometry
 Relationship between reactants and products
participated in chemical/biochemical reaction.
 Stoichiometric equation is a statement of the
relative number of molecules
 (or) moles of reactants and the products that
participate in the reaction.
 Ex: 2SO2+O2→2SO3
Biochemical Stoichiometry
Stoichiometric Coefficient:
The number before(in-front) of each molecule in
the Stoichiometric equation.
Stoichiometric Ratio:
Ratio of Stoichiometric coefficients between any
two chemical species in the Stoichiometric
equation.
Limiting reactant:-
 The reactant, that limits the amount of product formation
 That can be formed during the chemical reaction.
 The reaction will stop when the limiting reactant is consumed.
 The reactant which produce the less amount of product is the
limiting reactant.
 The reactant which ratio (moles/stoichiometric coefficient) is the
less, will be the limiting reactant
Excess reactant: -
 The reactant, that remains, when the reaction is
completed.
 The excess reactant remains because there is
noting with which it can react.
 The reactant which ratio (moles/stoichiometric
coefficient) is high, will be the excess reactant.
Stoichiometric ratio (SR):-
 The ratio of stoichiometric coefficients between
any two chemical species in stoichiometric
equation.
Biochemical Stoichiometry
Tutorial_3
200 moles of SO2 and 100 moles O2 are fed to a
reactor. Only 100 moles of SO2 react according to 𝑆
𝑂2 + 𝑂2 ⇌ 𝑆𝑂3.
Find the following
a) Fractional conversion
b) percentage conversion
c) Fraction unreacted
d) Stoichiometric ratio of SO2 to SO3
200 moles of SO2 and 100 moles O2 are fed to a
reactor. Only 100 moles of SO2 react according
to 𝑆𝑂2 + 𝑂2 ⇌ 𝑆𝑂3.
Given data:
Stoichiometric equation:
2SO2+O2→2SO3.
SO2 fed to reactor =200 moles
SO2 reacted =100 moles
O2 fed to reactor= 100 moles
200 moles of SO2 and 100 moles O2 are fed to a
reactor. Only 100 moles of SO2 react according
to 𝑆𝑂2 + 𝑂2 ⇌ 𝑆𝑂3.
Tutorial_4
A 2 grams sample of NH3 mixed with 4 grams of
oxygen presence of a suitable catalyst to form
NO and Water. Determine the limiting reactant.
𝑁𝐻3 + 𝑂2 ⇌ 𝑁𝑂 + 𝐻2𝑂
Given data
NH3= 2 g
O2= 4 g
Molecular weights
MW of NH3= 17
MW of O2= 32
MW of NO= 30
• Stoichiometric equation:
4𝑁𝐻3 + 5𝑂2 ⇌ 4𝑁𝑂 + 6𝐻2𝑂
Method 1:
In the case of NH3:
• (4x17) gr of NH3 produce→ (4x30) gr of NO
• Therefore, 1 gr of NH3 produce → (4x30)
(4x17) = 1.76 gr of NO
• For 2 gr of NH3 produce → 2 x1.74= 3.52 gr
of NO
In the case of O2:
• Similarly, (5x32) gr of O2 produce → (4x30) gr
of NO
• Therefore, 1 gr of O2 produce → (4x30) (4x17)
= 0.77 gr of NO
• For 4 gr of O2 produce → 4 x0.77= 3 gr of NO
Since, the reactant which produce the less amount of
product is the limiting reactant, therefore O2 is the
limiting reactant.
Since, the reactant which ratio
is the less, will be the limiting reactant, therefore O2 is the
limiting reactant.
Equilibrium Constant (K)
• The ratio between forward reaction rate constant
to the backward reaction rate constant when the
reaction is reversible/equilibrium is called
equilibrium constant (K).
• K=Forward reaction rate constant(k1) / Backward
reaction rate constant (k2).
• For example: A + B ⇌ C+ D
Equilibrium Constant (K)
• For example: A + B ⇌ C+ D
Equilibrium Constant (K)
Tutorial_5
Determine the fraction of pure Glucose that would
be converted to Pyruvate at 333 K and 1 atm, if the
following glycolysis reaction goes to equilibrium
during cellular synthesis process.
𝐶6𝐻12𝑂6 ⇌ 2𝐶3𝐻4𝑂3 + 2𝐻2
ΔG for the reaction at 333 K = 25 kJ. Assume ideal
gas behaviour for 1 mole.
Tutorial_5
Given data
ΔG =25 kJ/mole
T= 333 K
R= 8.314 kJ/mole. K
P=1 atm.
From the Equilibrium Constant (K) and the Standard
Gibbs Free Energy (ΔG) relation
ΔG= -RT ln K
Tutorial_5
So that, the fraction of pure glucose is converted to
pyruvate and hydrogen is approximately about 55%
Tutorial_5
Growth Stoichiometry and
Elemental Balances
Growth Stoichiometry and
Elemental Balances
Biomass Yield
Product Yield (YPS)
 YPS does not hold if product formation is not
directly linked with growth.
 Accordingly it cannot be applied for secondary-
metabolite production
 Ex: Penicillin fermentation, or for
biotransformation
 such as steroid hydroxylation which involve only
a small number of enzymes in cells.
Tutorial_6
Consider the following pair of reactions:
A→ B (Desired)
A→ C (Undesired),
200moles of A are fed to a batch reactor and the
final product contains 10mol of A, 150mol of B
and 10mol of C.
Determine the following:
a) Percentage yield of B
b) Selectivity of B relative to C
Tutorial_6
Given data:
Reactant(A) consumed: 200 moles
Desired product (B) formed= 150 moles
Undesired product (C) formed= 10 moles
Laws of Thermodynamics in
Biochemical Reactions
Biological Energy
Gibbs free energy (G): -
Biochemical Reactions
 The change in Gibbs Free Energy can be
used to determine whether a given chemical
reaction can occur spontaneously or not.
 If ΔG is negative, the reaction can occur
spontaneously.
 Likewise, if ΔG is positive, the reaction is
nonspontaneous.
Gibbs free energy (G): -
Biochemical Reactions
 Chemical reactions can be “coupled” together if
they share intermediates.
 In this case, the overall Gibbs Free Energy
change is simply the sum of the ΔG values for
each reaction.
 Therefore, an unfavourable reaction (positive
ΔG1) can be driven by a second, highly
favourable reaction (negative ΔG2 where the
magnitude of ΔG2 > magnitude of ΔG1).
Gibbs free energy (G): -
Biochemical Reactions
 For example, the reaction of glucose with
fructose to form sucrose has a ΔG value of +5.5
kcal/mole.
Glucose + Fructose  Sucrose ------(1)
ΔG = +5.5 kcal/mole.
 Therefore, this reaction will not occur
spontaneously.
Gibbs free energy (G): -
Biochemical Reactions
 The breakdown of ATP to form ADP and
inorganic phosphate has a ΔG value of -7.3
kcal/mole.
ATP  ADP + Pi -------(2)
ΔG = -7.3 kcal/mole.
 Therefore, this reaction will occur
spontaneously.
Gibbs free energy (G): -
Biochemical Reactions
 These two reactions can be coupled together, so
that glucose binds with ATP to form glucose-1-
phosphate and ADP.
Glucose+ ATP Glucose-1-phosphate +ADP
 The glucose-1-phosphate is then able to bond
with fructose yielding sucrose and inorganic
phosphate.
Glucose-1-phosphate +Fructose  Sucrose + Pi
Gibbs free energy (G): -
Biochemical Reactions
 The ΔG value of the coupled reaction is -1.8 kcal/mole,
indicating that the reaction will occur spontaneously.
ΔG = ΔG1+ ΔG2
= +5.5+ (-7.3)
= -1.8 kcal/mole.
ΔG<0 therefore, the reaction will become spontaneously.
 This principle of coupling reactions to alter the change
in Gibbs Free Energy is the basic principle behind all
enzymatic action in biological organisms.
Thank You

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Biochemical Thermodynamics for Biotechnology

  • 2. Overview of the Presentation  Biochemical Stoichiometry  Equilibrium Constant and Gibbs Free Energy  Growth Stoichiometry and Elemental Balance  Biomass Yield and Product Yield  Laws of Thermodynamics in Biochemical Reactions
  • 3. Biochemical Stoichiometry  Relationship between reactants and products participated in chemical/biochemical reaction.  Stoichiometric equation is a statement of the relative number of molecules  (or) moles of reactants and the products that participate in the reaction.  Ex: 2SO2+O2→2SO3
  • 4. Biochemical Stoichiometry Stoichiometric Coefficient: The number before(in-front) of each molecule in the Stoichiometric equation. Stoichiometric Ratio: Ratio of Stoichiometric coefficients between any two chemical species in the Stoichiometric equation.
  • 5. Limiting reactant:-  The reactant, that limits the amount of product formation  That can be formed during the chemical reaction.  The reaction will stop when the limiting reactant is consumed.  The reactant which produce the less amount of product is the limiting reactant.  The reactant which ratio (moles/stoichiometric coefficient) is the less, will be the limiting reactant
  • 6. Excess reactant: -  The reactant, that remains, when the reaction is completed.  The excess reactant remains because there is noting with which it can react.  The reactant which ratio (moles/stoichiometric coefficient) is high, will be the excess reactant. Stoichiometric ratio (SR):-  The ratio of stoichiometric coefficients between any two chemical species in stoichiometric equation.
  • 8. Tutorial_3 200 moles of SO2 and 100 moles O2 are fed to a reactor. Only 100 moles of SO2 react according to 𝑆 𝑂2 + 𝑂2 ⇌ 𝑆𝑂3. Find the following a) Fractional conversion b) percentage conversion c) Fraction unreacted d) Stoichiometric ratio of SO2 to SO3
  • 9. 200 moles of SO2 and 100 moles O2 are fed to a reactor. Only 100 moles of SO2 react according to 𝑆𝑂2 + 𝑂2 ⇌ 𝑆𝑂3. Given data: Stoichiometric equation: 2SO2+O2→2SO3. SO2 fed to reactor =200 moles SO2 reacted =100 moles O2 fed to reactor= 100 moles
  • 10. 200 moles of SO2 and 100 moles O2 are fed to a reactor. Only 100 moles of SO2 react according to 𝑆𝑂2 + 𝑂2 ⇌ 𝑆𝑂3.
  • 11. Tutorial_4 A 2 grams sample of NH3 mixed with 4 grams of oxygen presence of a suitable catalyst to form NO and Water. Determine the limiting reactant. 𝑁𝐻3 + 𝑂2 ⇌ 𝑁𝑂 + 𝐻2𝑂 Given data NH3= 2 g O2= 4 g Molecular weights MW of NH3= 17 MW of O2= 32 MW of NO= 30
  • 12. • Stoichiometric equation: 4𝑁𝐻3 + 5𝑂2 ⇌ 4𝑁𝑂 + 6𝐻2𝑂 Method 1: In the case of NH3: • (4x17) gr of NH3 produce→ (4x30) gr of NO • Therefore, 1 gr of NH3 produce → (4x30) (4x17) = 1.76 gr of NO • For 2 gr of NH3 produce → 2 x1.74= 3.52 gr of NO
  • 13. In the case of O2: • Similarly, (5x32) gr of O2 produce → (4x30) gr of NO • Therefore, 1 gr of O2 produce → (4x30) (4x17) = 0.77 gr of NO • For 4 gr of O2 produce → 4 x0.77= 3 gr of NO Since, the reactant which produce the less amount of product is the limiting reactant, therefore O2 is the limiting reactant.
  • 14. Since, the reactant which ratio is the less, will be the limiting reactant, therefore O2 is the limiting reactant.
  • 15. Equilibrium Constant (K) • The ratio between forward reaction rate constant to the backward reaction rate constant when the reaction is reversible/equilibrium is called equilibrium constant (K). • K=Forward reaction rate constant(k1) / Backward reaction rate constant (k2). • For example: A + B ⇌ C+ D
  • 16. Equilibrium Constant (K) • For example: A + B ⇌ C+ D
  • 18. Tutorial_5 Determine the fraction of pure Glucose that would be converted to Pyruvate at 333 K and 1 atm, if the following glycolysis reaction goes to equilibrium during cellular synthesis process. 𝐶6𝐻12𝑂6 ⇌ 2𝐶3𝐻4𝑂3 + 2𝐻2 ΔG for the reaction at 333 K = 25 kJ. Assume ideal gas behaviour for 1 mole.
  • 19. Tutorial_5 Given data ΔG =25 kJ/mole T= 333 K R= 8.314 kJ/mole. K P=1 atm. From the Equilibrium Constant (K) and the Standard Gibbs Free Energy (ΔG) relation ΔG= -RT ln K
  • 21. So that, the fraction of pure glucose is converted to pyruvate and hydrogen is approximately about 55% Tutorial_5
  • 25. Product Yield (YPS)  YPS does not hold if product formation is not directly linked with growth.  Accordingly it cannot be applied for secondary- metabolite production  Ex: Penicillin fermentation, or for biotransformation  such as steroid hydroxylation which involve only a small number of enzymes in cells.
  • 26. Tutorial_6 Consider the following pair of reactions: A→ B (Desired) A→ C (Undesired), 200moles of A are fed to a batch reactor and the final product contains 10mol of A, 150mol of B and 10mol of C. Determine the following: a) Percentage yield of B b) Selectivity of B relative to C
  • 27. Tutorial_6 Given data: Reactant(A) consumed: 200 moles Desired product (B) formed= 150 moles Undesired product (C) formed= 10 moles
  • 28. Laws of Thermodynamics in Biochemical Reactions
  • 29.
  • 31.
  • 32.
  • 33.
  • 34.
  • 35.
  • 36. Gibbs free energy (G): - Biochemical Reactions  The change in Gibbs Free Energy can be used to determine whether a given chemical reaction can occur spontaneously or not.  If ΔG is negative, the reaction can occur spontaneously.  Likewise, if ΔG is positive, the reaction is nonspontaneous.
  • 37. Gibbs free energy (G): - Biochemical Reactions  Chemical reactions can be “coupled” together if they share intermediates.  In this case, the overall Gibbs Free Energy change is simply the sum of the ΔG values for each reaction.  Therefore, an unfavourable reaction (positive ΔG1) can be driven by a second, highly favourable reaction (negative ΔG2 where the magnitude of ΔG2 > magnitude of ΔG1).
  • 38. Gibbs free energy (G): - Biochemical Reactions  For example, the reaction of glucose with fructose to form sucrose has a ΔG value of +5.5 kcal/mole. Glucose + Fructose  Sucrose ------(1) ΔG = +5.5 kcal/mole.  Therefore, this reaction will not occur spontaneously.
  • 39. Gibbs free energy (G): - Biochemical Reactions  The breakdown of ATP to form ADP and inorganic phosphate has a ΔG value of -7.3 kcal/mole. ATP  ADP + Pi -------(2) ΔG = -7.3 kcal/mole.  Therefore, this reaction will occur spontaneously.
  • 40. Gibbs free energy (G): - Biochemical Reactions  These two reactions can be coupled together, so that glucose binds with ATP to form glucose-1- phosphate and ADP. Glucose+ ATP Glucose-1-phosphate +ADP  The glucose-1-phosphate is then able to bond with fructose yielding sucrose and inorganic phosphate. Glucose-1-phosphate +Fructose  Sucrose + Pi
  • 41. Gibbs free energy (G): - Biochemical Reactions  The ΔG value of the coupled reaction is -1.8 kcal/mole, indicating that the reaction will occur spontaneously. ΔG = ΔG1+ ΔG2 = +5.5+ (-7.3) = -1.8 kcal/mole. ΔG<0 therefore, the reaction will become spontaneously.  This principle of coupling reactions to alter the change in Gibbs Free Energy is the basic principle behind all enzymatic action in biological organisms.