1.2 Chemical Foundations

Step 1: Define Gibbs free energy from enthalpy and entropy

The Gibbs free energy G is defined as a state function combining enthalpy H and entropy S at constant temperature and pressure:

$$G = H - TS$$

Step 2: For a process at constant T and P, take the change

$$\Delta G = \Delta H - T\Delta S$$

This is the master equation: a reaction is spontaneous when $\Delta G < 0$, which can occur if $\Delta H < 0$ (exothermic) or $T\Delta S > 0$ (entropy increase), or both.

Step 3: Chemical potential and reaction quotient

For a reaction with concentrations differing from standard state, the free energy depends on the reaction quotient Q:

$$\Delta G = \Delta G° + RT\ln Q$$

where $Q = \frac{[\text{Products}]}{[\text{Reactants}]}$ and R = 8.314 J/(mol K).

Step 4: At equilibrium, $\Delta G = 0$ and $Q = K_{eq}$

$$0 = \Delta G° + RT\ln K_{eq}$$

Step 5: Solve for the standard free energy

$$\boxed{\Delta G° = -RT\ln K_{eq}}$$

This is the bridge between thermodynamics and equilibrium. A large $K_{eq}$ (products favored) corresponds to a large negative $\Delta G°$. For example, ATP hydrolysis: $K_{eq} \approx 2 \times 10^5$ gives $\Delta G°' = -30.5$ kJ/mol at 37 C.

Step 6: Numerical example at body temperature (T = 310 K)

$$\Delta G° = -(8.314 \times 10^{-3})(310)\ln(2 \times 10^5) = -2.578 \times \ln(2 \times 10^5) \approx -31.4 \text{ kJ/mol}$$

Step 1: The fundamental postulate of statistical mechanics

For an isolated system with total energy $E_{total}$, all accessible microstates are equally probable. Consider a small subsystem (a single molecule) in contact with a large heat bath (the rest of the cell).

Step 2: Probability is proportional to the number of reservoir states

When the molecule is in state i with energy $E_i$, the reservoir has energy $E_{total} - E_i$. The number of reservoir microstates is $\Omega(E_{total} - E_i)$, so:

$$P(E_i) \propto \Omega(E_{total} - E_i)$$

Step 3: Use the entropy-microstate relation and Taylor expand

Since $S = k_B \ln \Omega$, we get $\Omega = e^{S/k_B}$. Taylor-expanding the reservoir entropy around $E_{total}$:

$$S(E_{total} - E_i) \approx S(E_{total}) - E_i \frac{\partial S}{\partial E} = S(E_{total}) - \frac{E_i}{T}$$

where we used $\partial S/\partial E = 1/T$ (the thermodynamic definition of temperature).

Step 4: Exponentiate to get the Boltzmann factor

$$P(E_i) \propto e^{S(E_{total})/k_B} \cdot e^{-E_i/k_BT} \propto e^{-E_i/k_BT}$$

The first factor is a constant (same for all states) and is absorbed into normalization.

Step 5: Define the partition function Z for normalization

Since probabilities must sum to 1:

$$Z = \sum_i e^{-E_i/k_BT}$$

$$\boxed{P(E_i) = \frac{e^{-E_i/k_BT}}{Z}}$$

Step 6: Connection to free energy and biological significance

The Helmholtz free energy is $F = -k_BT \ln Z$. At room temperature ($T = 300$ K), $k_BT \approx 2.5$ kJ/mol $\approx 0.6$ kcal/mol. This sets the energy scale for non-covalent interactions: a hydrogen bond (~10 kJ/mol $\approx 4k_BT$) significantly shifts the population, while thermal fluctuations can break individual weak bonds, enabling dynamic molecular recognition.

Step 1: Start with the harmonic approximation near equilibrium

Near the equilibrium bond length $r_0$, Hooke's law gives:

$$V_{\text{harmonic}}(r) = \frac{1}{2}k(r - r_0)^2$$

This parabola goes to infinity at large r, which is unphysical -- real bonds dissociate at finite energy.

Step 2: Require correct asymptotic behavior

A realistic potential must satisfy: $V(r_0) = 0$ (minimum), $V(r \to \infty) = D_e$ (dissociation energy), and $V'(r_0) = 0$ (equilibrium). The Morse potential achieves this with an exponential form:

$$\boxed{E(r) = D_e\left(1 - e^{-a(r - r_0)}\right)^2}$$

Step 3: Verify the boundary conditions

At $r = r_0$: $E = D_e(1 - 1)^2 = 0$ (minimum). As $r \to \infty$: $e^{-a(r-r_0)} \to 0$, so $E \to D_e$ (dissociation plateau). As $r \to 0$: $E$ rises steeply (repulsion).

Step 4: Relate the width parameter a to the force constant k

Expanding around $r_0$ using $e^{-x} \approx 1 - x + x^2/2$:

$$E(r) \approx D_e \cdot a^2(r - r_0)^2 + \cdots$$

Comparing with the harmonic potential: $k = 2D_e a^2$, therefore:

$$a = \sqrt{\frac{k}{2D_e}}$$

Step 5: The force and its relationship to bond stretching

$$F(r) = -\frac{dE}{dr} = 2aD_e\left(1 - e^{-a(r-r_0)}\right)e^{-a(r-r_0)}$$

The maximum restoring force occurs at the inflection point. Beyond this, the bond weakens and eventually breaks.

Step 6: Biological application -- C-C bond parameters

For a C-C single bond: $D_e \approx 348$ kJ/mol, $r_0 = 0.154$ nm, $a \approx 19.0$ nm$^{-1}$. For C=C double bond: $D_e \approx 614$ kJ/mol, $r_0 = 0.134$ nm. The higher $D_e$ and shorter $r_0$ for double bonds reflects the general trend: stronger bonds are shorter, a consequence of greater orbital overlap.