Bioenergetics &
Thermodynamics
Master the principles of free energy, entropy, and enthalpy that govern every biochemical reaction in living systems.
MCAT Biochemistry
What Is Bioenergetics?
Bioenergetics is the study of energy flow through living systems. Every cellular process, from muscle contraction to DNA replication, requires energy transformations that obey the laws of thermodynamics.
The Three Key Players
\(\Delta G\): Gibbs Free Energy
The energy available to do useful work. Determines spontaneity of reactions.
\(\Delta H\): Enthalpy
Total heat content. Reflects bond energies made and broken during a reaction.
\(\Delta S\): Entropy
Measure of disorder. The universe always trends toward increasing entropy.
Laws of Thermodynamics
1
First Law (Conservation): Energy cannot be created or destroyed, only converted from one form to another. Total energy of an isolated system is constant.
2
Second Law (Entropy): In any spontaneous process, the total entropy of the universe increases. No process is 100% efficient: some energy is always lost as heat.
The Gibbs Free Energy Equation
\(\Delta G\)
Change in free energy (kJ/mol)
\(\Delta H\)
Change in enthalpy (kJ/mol)
\(T\)
Temperature in Kelvin (K)
\(\Delta S\)
Change in entropy (J/mol·K)
Spontaneity: Exergonic vs Endergonic
\(\Delta G < 0\)
Exergonic: Spontaneous, releases free energy. Products are more stable.
\(\Delta G > 0\)
Endergonic: Non-spontaneous, requires energy input. Driven by coupling.
Exothermic vs Endothermic
| Exothermic | Endothermic | |
|---|---|---|
| \(\Delta H\) | Negative (\(< 0\)) | Positive (\(> 0\)) |
| Heat | Released to surroundings | Absorbed from surroundings |
| Bond Energy | Products have stronger bonds | Reactants have stronger bonds |
| Example | Combustion of glucose | Photosynthesis |
Coupling Reactions and ATP
Cells drive endergonic reactions by coupling them to the hydrolysis of ATP, a highly exergonic reaction:
\(\Delta G^\circ\) vs \(\Delta G\)
\(\Delta G^\circ\) Standard
Measured at standard conditions: \(25^\circ\text{C}\), 1 atm, 1 M concentrations, pH 7 (for biochemistry: \(\Delta G^{\circ\prime}\)).
\(\Delta G\) Actual
Depends on actual concentrations and conditions inside the cell. This is what determines real spontaneity.
Temperature and Spontaneity
| \(\Delta H\) | \(\Delta S\) | \(\Delta G\) | Spontaneous? |
|---|---|---|---|
| \(< 0\) | \(> 0\) | Always \(< 0\) | Always |
| \(> 0\) | \(< 0\) | Always \(> 0\) | Never |
| \(< 0\) | \(< 0\) | Depends on T | Low T favors |
| \(> 0\) | \(> 0\) | Depends on T | High T favors |
How to Predict Spontaneity
1
Identify \(\Delta H\): Is the reaction exothermic (bonds formed > bonds broken) or endothermic?
2
Identify \(\Delta S\): Does disorder increase (more gas molecules, dissolution) or decrease (crystallization, fewer molecules)?
3
Apply \(\Delta G = \Delta H - T\Delta S\): Calculate or reason whether the result is negative (spontaneous) or positive (non-spontaneous).
4
Consider conditions: Check if actual concentrations (\(Q\)) or coupling to ATP shifts the effective \(\Delta G\).
Example 1: Glucose Combustion
1
Convert T to Kelvin: \(T = 37 + 273 = 310\) K
2
Convert \(\Delta S\) to kJ: \(\Delta S = 182.4 \text{ J/mol·K} = 0.1824 \text{ kJ/mol·K}\)
3
Calculate \(T\Delta S\): \(310 \times 0.1824 = 56.5 \text{ kJ/mol}\)
4
Calculate \(\Delta G\): \(\Delta G = -2803 - 56.5 = \) \(-2859.5\) kJ/mol
Example 2: Protein Folding
1
Set \(\Delta G = 0\): At equilibrium, \(0 = \Delta H - T\Delta S\)
2
Solve for T: \(T = \frac{\Delta H}{\Delta S} = \frac{-80{,}000}{-200} = 400\) K = \(127^\circ\text{C}\)
3
Interpret: Below 400 K: \(\Delta G < 0\) (spontaneous). Above 400 K: \(\Delta G > 0\) (non-spontaneous, protein denatures).
Example 3: \(\Delta G\) and Equilibrium
1
Rearrange: \(\ln K_{eq} = \frac{-\Delta G^\circ}{RT} = \frac{17{,}100}{8.314 \times 298} = 6.90\)
2
Solve: \(K_{eq} = e^{6.90} \approx\) 1000
Common Mistakes
✕ Confusing \(\Delta G\) with \(\Delta H\)
Exothermic does NOT always mean spontaneous. You must consider the entropy term \(T\Delta S\) as well.
✕ Unit Mismatch
\(\Delta H\) is often in kJ/mol while \(\Delta S\) is in J/mol·K. Always convert to the same unit before calculating.
✕ Using \(\Delta G^\circ\) for Real Cells
Standard \(\Delta G^\circ\) is measured at 1 M concentrations. Actual cellular \(\Delta G\) can be very different due to low concentrations.
Summary of Key Relationships
| Concept | Equation | What It Tells You |
|---|---|---|
| Free Energy | \(\Delta G = \Delta H - T\Delta S\) | Spontaneity of a reaction |
| Actual \(\Delta G\) | \(\Delta G = \Delta G^\circ + RT\ln Q\) | Effect of concentrations |
| Equilibrium | \(\Delta G^\circ = -RT\ln K_{eq}\) | Position of equilibrium |
| At Equilibrium | \(\Delta G = 0\) | No net reaction occurs |
| Coupling | \(\Delta G_{net} = \Sigma \Delta G_i\) | Additive free energies |
Question 1 of 5
A reaction has \(\Delta H = -40\) kJ/mol and \(\Delta S = +100\) J/mol·K. At 298 K, which of the following is true?
When \(\Delta H < 0\) and \(\Delta S > 0\), both terms make \(\Delta G\) negative: \(\Delta G = \Delta H - T\Delta S = (-40) - (298)(0.1) = -69.8\) kJ/mol. This combination is always spontaneous regardless of temperature.
Question 2 of 5
The hydrolysis of ATP has \(\Delta G^{\circ\prime} = -30.5\) kJ/mol. An endergonic reaction requires +18 kJ/mol. What is the \(\Delta G\) of the coupled reaction?
Free energies are additive in coupled reactions: \(\Delta G_{net} = (-30.5) + (+18) = -12.5\) kJ/mol. The net negative value means the coupled process is spontaneous.
Question 3 of 5
At equilibrium, which of the following must be true?
At equilibrium, the actual \(\Delta G = 0\) (no net driving force). However, \(\Delta G^\circ\) is a fixed property of the reaction and is typically not zero. They are related by \(\Delta G^\circ = -RT\ln K_{eq}\).
Question 4 of 5
A reaction has \(\Delta H > 0\) and \(\Delta S > 0\). Under what conditions is it spontaneous?
When both \(\Delta H\) and \(\Delta S\) are positive, \(\Delta G = \Delta H - T\Delta S\) becomes negative only when \(T\Delta S > \Delta H\). This requires high temperature to make the entropy term large enough to overcome the positive enthalpy.
Question 5 of 5
According to the Second Law of Thermodynamics, how do living organisms maintain their highly ordered structure?
Living organisms are open systems that exchange matter and energy with their surroundings. They maintain low internal entropy by releasing heat and waste products, which increases the entropy of the surroundings by a greater amount. The total entropy of the universe still increases.