L2

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Bio98 Lecture 2

  • Topic: Bioenergetics

  • Section: 1.4

Simple Thermodynamics

  • Concept of Spontaneity:

    • A process is considered spontaneous if the reaction moving to the right is favored.

    • For spontaneous reactions: K_{eq} > 1.0

  • Equilibrium Constant (Keq):

    • Defined as: Keq=rac[products][reactants]K_{eq} = rac{[products]}{[reactants]}

    • For the general reaction: aA+bB<br>ightleftarrowscC+dDaA + bB <br>ightleftarrows cC + dD, it can be expressed as:
      Keq=rac[C]c[D]d[A]a[B]bK_{eq} = rac{[C]^c[D]^d}{[A]^a[B]^b}

  • Gibbs Free Energy (G):

    • Condition for spontaneity:

    • If G^ ext{o}{reactants} > G^ ext{o}{products}, then G^ ext{o}_{reaction} < 0

    • Standard Gibbs Free Energy equation:
      G = G^ ext{o} + RT ext{ln} rac{[C]^c[D]^d}{[A]^a[B]^b}
      where:

    • TT is the temperature,

    • RR is the universal gas constant,

    • [A],[B],[C],[D][A], [B], [C], [D] are the concentrations of reactants/products.

  • Change in Gibbs Free Energy at various conditions:

    • High concentration of products makes G more positive (or less negative).

    • High concentration of reactants makes G more negative (or less positive).

    • At equilibrium:

    • G = 0 and Q=KeqQ = K_{eq}

    • Relation:
      K_{eq} = e^{- rac{G^ ext{o}}{RT}}

Example: Interconversion of Sugars

  • Significant biological reaction:

    • Interconversion:

    • Glucose-6-phosphate (G6P) and Fructose-6-phosphate (F6P).

Concentration Calculations

  • Suppose you dissolve:

    • 1 mole of G6P and 0.3 moles of F6P into a 1L solution.

  • Equilibrium Concentrations:

    • [F6P]=0.44M[F6P] = 0.44M

    • Total concentration: F6P+G6P=1.3MF6P + G6P = 1.3M

    • To find concentration of G6P:

    • G6P=1.3M−F6P=1.3M−0.44M=0.86MG6P = 1.3M - F6P = 1.3M - 0.44M = 0.86M

  • Equilibrium Constant (Keq):

    • Keq=rac[F6P][G6P][1.3M−F6P]=0.504K_{eq} = rac{[F6P][G6P]}{[1.3M - F6P]} = 0.504

Further Evaluation of Reaction Favorability

  • Conclusion:

    • Reaction is not favored as written:

    • Calculate KeqK_{eq} at room temperature:

    • Given values for calculations:

      • R=8.314extJK−1extmol−1R = 8.314 ext{ J K}^{-1} ext{mol}^{-1}

      • K_{eq} = e^{- rac{G^ ext{o}}{RT}} = e^{- rac{-14,000 ext{ J/mol}}{8.314 ext{ J K}^{-1} ext{mol}^{-1} imes 298 ext{ K}}}

    • Equilibrium of the reaction involving G6P:

    • G6P+HPO4−−<spandata−name="leftrightarrow"data−type="emoji">↔</span>Glu−6−phosphateG6P + HPO4^{--} <span data-name="left_right_arrow" data-type="emoji">↔</span> Glu-6-phosphate

      • G^ ext{o} = +14 kJ/mol indicates favorability

Energy Coupling

  • Definition: Enzymes can utilize the energy from a favorable reaction to drive an unfavorable reaction.

  • Example Reactions:

    1. Reaction (1):

    • A o B
      ightarrow G^ ext{o} = +20 kJ/mol (unfavorable)

    1. Reaction (2):

    • C o D
      ightarrow G^ ext{o} = -31 kJ/mol (favorable)

  • Net Reaction:

    • Combining (1) and (2):
      A + C o B + D
      ightarrow G^ ext{o} = -11 kJ/mol (favorable).

  • Specific Example: Hexokinase reaction

    • Glucose + ATP → G-6-P + ADP

    • G^ ext{o} = -17 ext{ kJ/mol} (reaction favored)

    • Keq=955K_{eq} = 955 at T=298KT = 298 K

Context for Proton Release and ATP Hydrolysis

  • Proton Release:

    • Intracellular Concentration: [H+] is maintained very low at approximately 10−7M10^{-7} M which favors reactions that release protons.

  • Why is ATP Hydrolysis Highly Exergonic?:

    1. Resonance Forms: More stability in the products

    2. Hydration Stabilization: Stabilization is more favorable in products

    3. Electrostatic Repulsion: Less in products.

  • Standard Gibbs Free Energy for ATP hydrolysis:

    • G^ ext{o} = -31 kJ/mol

Fundamental Thermodynamic Equation

  • G = H - TS

    • Where:

    • H = change in enthalpy

    • S = change in entropy

  • Enthalpy Definition:

    • H = H{products} - H{reactants}

    • Reaction is favorable when H < 0

Entropy and Enthalpy Examples

  • Example Reaction:

    • N2O5(s)o2NO2(g)+rac12O2(g)N2O5 (s) o 2 NO2 (g) + rac{1}{2} O2 (g)

    • G^ ext{o} = -31.2 ext{ kJ/mol}, H^ ext{o} = +109.0 ext{ kJ/mol}

  • Determining Entropy Change:

    • Use:
      G^{ ext{o}} = H^{ ext{o}} - TS^{ ext{o}}

    • Applying known values, find S^{ ext{o}} :

    • Calculation yields:
      S^{ ext{o}} = +0.47 ext{ kJ mol}^{-1} K^{-1}

    • Result indicates that the reaction is entropically favorable.

Exothermic Reaction Example

  • Classic Example: Reaction of hydrogen and oxygen forming water:

    • Reaction formula:
      O2+2H2o2H2OO2 + 2 H2 o 2 H2O

    • Notable properties:

    • High exothermic nature—release of heat.

    • Phenomenon when hydrogen filled balloon is ignited, a rapid and intense reaction ensures (explosion).

Acid-Base Equilibria, pH, and Buffers

  • Sections: 2.2 - 2.3

pH Profiles of Enzymatic Reactions

  • Example Enzymes and Activity across pH:

    • Salivary Amylase (1-10): Active at around neutral pH.

    • Pepsin: Optimal at acidic pH.

  • General Observation: Enzyme activity varies significantly across different pH levels.

Water Ionization and pH

  • Pure Water: Slightly ionized:

    • Dissociation of Water:

    • H2O<br>ightleftarrowsH++OH−H2O <br>ightleftarrows H^+ + OH^-

  • Hydronium Formation:

    • Free protons do not exist alone in solutions; they exist as hydronium ions (H3O+).

Achieving Equilibrium in Water

  • Ion Product of Water (Kw):

    • Defined by:
      Kw=[H+][OH−]=[H+]2=1.0imes10−14M2K_w = [H^+][OH^-] = [H^+]^2 = 1.0 imes 10^{-14} M^2

    • At neutrality: [H^+] = [OH^-] = 10^{-7} M = 0.1 4M

    • Water concentration in a solution:

    • Pure water has equal quantities of [H+][H^+] and [OH−][OH^-].

  • Concentration Relationship:

    • K<em>w=K</em>eqimes[H2O]K<em>w = K</em>{eq} imes [H2O]; indicating the constant nature of water ion products.

pH Scale Basics

  • General pH Equation:

    • pH=−extlog[H+]pH = - ext{log}[H^+]

    • Example for 1M HCl:

      • pH=−extlog(1)=0pH = - ext{log}(1) = 0

  • Understanding Strong Acids and Bases:

    • Strong Acids: Completely dissociate in solution (e.g., HCl yields [H+] also at 1 M).

    • Strong Bases: Completely dissociate in solution (e.g., NaOH gives [OH-] at 1M leading to [H+] at 10−14M10^{-14} M resulting in a pH of 14).

Dissociation of Weak Acids and Bases

  • Weak Acid Dissociation Equation: Ka=rac[H+][A−][HA]Ka = rac{[H^+][A^-]}{[HA]}

    • Derived from the reaction:

    • R−C−OH<br>ightleftarrowsR−C−O−+H+R-C-OH <br>ightleftarrows R-C-O^- + H^+

    • R−NH3+<br>ightleftarrowsR−NH2+H+R-NH3^+ <br>ightleftarrows R-NH2 + H^+

Example of Weak Acid Acidity Constant (Ka)

  • Example Calculation:

    • For acetic acid, pK<em>a=−extlog(K</em>a)pK<em>a = - ext{log}(K</em>a)

  • Given: Ka=1.76imes10−5MK_a = 1.76 imes 10^{-5} M,

    • Then,

    • pKa=4.8pK_a = 4.8 (measure of strength of this weak acid).

Henderson-Hasselbalch Equation

  • Equation:

    • pH=pKa+extlograc[A−][HA]pH = pK_a + ext{log} rac{[A^-]}{[HA]}

    • Derivation using logarithmic properties from the dissociation constants.

Application Example of Weak Acid Dissociation in pH Calculations

  • Situation: Calculating for weak acetic acid:

    • Starting with 0.10M acetic acid and deducing dissociation to determine potential pH shifts across added hydroxide.

    • Approximating changes across varying ratios of conjugate acid/base.

    • Resulting dynamics illustrate the counteracting nature of the equilibrium equation as [H+] is altered due to reactions at endpoint adjustments of titration.