7.2. Equilibrium Constants from Microscopic Properties

Overview and Learning Objectives¶

Section 7.1 derived the equilibrium constant from macroscopic thermochemistry: given $\Delta_r H^\circ$ and $\Delta_r S^\circ$ (e.g., from NIST/JANAF/ATcT tables), one computes $\Delta_r G^\circ$ and then $K = \exp(-\Delta_r G^\circ/RT)$ .

In this section we derive the same $K$ from microscopic information — molecular masses, rotational and vibrational constants, electronic degeneracies, and bond dissociation energies. This completes the statistical-mechanical bridge built in Chapters 2, 4, and 5: Chapter 4 connected the canonical partition function to the Helmholtz free energy, $A = -k_{\mathrm B}T\ln Q$ ; Chapter 5 showed that $A$ is the master potential at constant $T,V$ , generating $S$ , $U$ , $P$ , and — as we now develop — the chemical potentials $\mu_i$ . Once we have $\mu_i$ in terms of the single-molecule partition functions $q_i$ , the equilibrium condition $\sum_i\nu_i\mu_i = 0$ gives $K$ directly in terms of molecular properties.

The payoff: for any gas-phase reaction, we can predict $K(T)$ from spectroscopic data alone, and we can see which molecular features (masses, moments of inertia, bond strengths) drive the equilibrium.

Learning objectives:

Write the equilibrium condition at fixed $T,V$ as $\sum_i \nu_i\mu_i=0$ from minimizing the Helmholtz free energy.
Use $Q = \prod_i q_i^{N_i}/N_i!$ for an ideal-gas mixture to derive $\mu_i$ from $A = -k_{\mathrm B}T\ln Q$ .
Express equilibrium constants in terms of single-molecule partition functions, and relate $K_p$ and $K_c$ through $\Delta\nu$ .
Identify how translational, rotational, vibrational, and electronic factors contribute to $K$ for a specific reaction, and compute $K(T)$ numerically from molecular constants.

Core Ideas and Derivations¶

7.2.1 The right potential for equilibrium at constant $T,V$ ¶

Consider a general gas-phase reaction

\nu_A A(g) + \nu_B B(g) \rightleftharpoons \nu_Y Y(g) + \nu_Z Z(g),

(1)

under the ideal-gas assumption used throughout this chapter.

Section 5.1 established that at constant $T$ and $V$ , the relevant thermodynamic potential is the Helmholtz free energy $A$ , with differential

dA = -S\,dT - P\,dV + \sum_i \mu_i\,dn_i.

(2)

At constant $T$ and $V$ ,

dA = \sum_i \mu_i\,dn_i = \left(\sum_i \nu_i\,\mu_i\right)d\xi,

(3)

where the second equality uses $dn_i = \nu_i\,d\xi$ from § 7.1.1. Equilibrium at fixed $T,V$ occurs at a minimum of $A$ , so

\left(\frac{\partial A}{\partial \xi}\right)_{T,V} = 0 \qquad\Longrightarrow\qquad \boxed{\sum_i \nu_i\,\mu_i = 0.}

(4)

This is the chemical equilibrium condition written in terms of chemical potentials. It is the same condition as in § 7.1 — the equilibrium position of a reaction does not depend on whether we hold $P$ or $V$ fixed, because $\mu_i$ is intensive. What changes between the two cases is only which potential is stationary: $G$ at fixed $T,P$ , $A$ at fixed $T,V$ .

7.2.2 Partition function of an ideal-gas mixture¶

From § 2.4 (factorization over independent subsystems) and § 2.5 (indistinguishable particles), the canonical partition function of a mixture of ideal gases factorizes:

\boxed{ Q(T,V,\{N_i\}) = \prod_i Q_i(T,V,N_i), \qquad Q_i(T,V,N_i) = \frac{q_i(T,V)^{N_i}}{N_i!}, }

(5)

where $q_i$ is the single-molecule partition function for species $i$ . Each species is treated independently because ideal-gas molecules don’t interact.

7.2.3 Chemical potentials from the partition function¶

Section 4.3 established

A = -k_{\mathrm B}T\ln Q,

(6)

and § 5.1 developed $A$ as the “master potential” at fixed $T,V$ , with $S$ , $U$ , $P$ all following from its derivatives. The chemical potential $\mu_i$ is the natural extension of that derivative machinery to composition:

\mu_i^{\text{(per molecule)}} \equiv \left(\frac{\partial A}{\partial N_i}\right)_{T,V,\{N_{j\neq i}\}} = -k_{\mathrm B}T\left(\frac{\partial \ln Q}{\partial N_i}\right)_{T,V,\{N_{j\neq i}\}}.

(7)

Using Eq. (5),

\ln Q = \sum_i \left(N_i\ln q_i - \ln N_i!\right),

(8)

and applying Stirling’s approximation in the thermodynamic limit ( $\ln N_i! \approx N_i\ln N_i - N_i$ , as introduced in § 2.5),

\left(\frac{\partial \ln Q}{\partial N_i}\right)_{T,V,\{N_{j\neq i}\}} = \ln q_i - \ln N_i = \ln\!\left(\frac{q_i}{N_i}\right).

(9)

So the per-molecule chemical potential is

\mu_i^{\text{(per molecule)}} = -k_{\mathrm B}T\ln\!\left(\frac{q_i}{N_i}\right).

(10)

Multiplying by Avogadro’s number $N_{\mathrm A}$ to express $\mu_i$ on a per-mole basis (as used throughout § 7.1),

\boxed{ \mu_i = -RT\ln\!\left(\frac{q_i}{N_i}\right) = RT\ln\!\left(\frac{N_i}{q_i}\right), }

(11)

where $R = N_{\mathrm A}k_{\mathrm B}$ . Equation (11) is the microscopic counterpart of the thermodynamic expression $\mu_i = \mu_i^\circ(T) + RT\ln(P_i/P^\circ)$ from § 7.1.3: both express the chemical potential of a gas-phase species, one in terms of partial pressure and a reference $\mu_i^\circ(T)$ , the other in terms of microscopic counting through $q_i/N_i$ .

7.2.4 Equilibrium constants from partition functions¶

Inserting Eq. (11) into the equilibrium condition $\sum_i \nu_i\mu_i=0$ :

\sum_i \nu_i\ln\!\left(\frac{q_i}{N_i}\right)=0 \qquad\Longrightarrow\qquad \prod_i\left(\frac{q_i}{N_i}\right)^{\nu_i}=1.

(12)

Equivalently,

\boxed{ \prod_i N_i^{\nu_i} = \prod_i q_i^{\nu_i}. }

(13)

Relating to $K_c$ and $K_p$ ¶

Define a number concentration $c_i = N_i/V$ . Dividing Eq. (13) by $V^{\Delta\nu}$ (with $\Delta\nu \equiv \sum_i \nu_i$ ) gives

K_c \equiv \prod_i c_i^{\nu_i} = \prod_i\!\left(\frac{q_i}{V}\right)^{\nu_i}.

(14)

For an ideal gas, $P_i = c_i\, k_{\mathrm B}T$ , so the pressure-based equilibrium constant is

\boxed{ K_p \equiv \prod_i\left(\frac{P_i}{P^\circ}\right)^{\nu_i} = K_c\,\left(\frac{k_{\mathrm B}T}{P^\circ}\right)^{\Delta\nu}. }

(15)

The prefactor $(k_{\mathrm B}T/P^\circ)^{\Delta\nu}$ converts between “molecules per unit volume” and “pressure ratios relative to $P^\circ$ .” When $\Delta\nu = 0$ , this prefactor is unity and $K_p = K_c$ is dimensionless directly.

So, once you can compute the single-molecule partition functions $q_i$ , you can compute $K_c$ (and then $K_p$ ).

7.2.5 Example: $\mathrm{HI(g)}$ formation¶

Consider

\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}.

(16)

Here,

\Delta\nu = 2 - 1 - 1 = 0,

(17)

so $K_p = K_c$ and the equilibrium constant is

\boxed{ K = \frac{q_{HI}^2}{q_{H_2}\,q_{I_2}}. }

(18)

The $\Delta\nu = 0$ choice is pedagogically convenient: it makes volume factors cancel exactly (see the Worked Example below), so $K$ depends only on intrinsic molecular properties (masses, rotational and vibrational constants, bond strengths). With $\Delta\nu \neq 0$ reactions, one would need the $(k_{\mathrm B}T/P^\circ)^{\Delta\nu}$ factor from Eq. (15) to convert between $K_c$ and $K_p$ .

7.2.6 Single-molecule partition function for a diatomic ideal gas¶

Within the rigid-rotor/harmonic-oscillator approximations developed in Chapter 2 (see § 2.8 for the summary table), the single-molecule partition function for a diatomic gas factors as

q = q_{\mathrm{trans}}\,q_{\mathrm{rot}}\,q_{\mathrm{vib}}\,q_{\mathrm{elec}}.

(19)

Energy-zero convention

A partition function depends on where one places the zero of energy. In this section we choose the zero of molecular energy to lie at the dissociated atoms at rest — i.e., the ground vibrational level of a bound diatomic sits at energy $-D_0$ , where $D_0$ is the bond dissociation energy measured from $v=0$ . This choice puts the bond energy directly in the electronic factor as $e^{+D_0/k_{\mathrm B}T}$ (per molecule) or $e^{+D_0/RT}$ (per mole), and keeps the vibrational factor in its “ground-state-at-zero” form $e^{-\Theta_{\mathrm{vib}}/(2T)}/(1-e^{-\Theta_{\mathrm{vib}}/T})$ . Some texts instead absorb the $v=0$ zero-point energy into the electronic factor and use $q_{\mathrm{vib}} = 1/(1-e^{-\Theta_{\mathrm{vib}}/T})$ ; both conventions give the same $K$ provided they are used consistently. We use the first convention throughout.

Putting the four factors together (cf. § 2.8):

\boxed{ q_{\text{diatomic}} = \underbrace{\frac{V}{\Lambda^3}}_{q_{\mathrm{trans}}} \;\underbrace{\frac{T}{\sigma\,\Theta_{\mathrm{rot}}}}_{q_{\mathrm{rot}}} \;\underbrace{\frac{e^{-\Theta_{\mathrm{vib}}/(2T)}}{1-e^{-\Theta_{\mathrm{vib}}/T}}}_{q_{\mathrm{vib}}} \;\underbrace{g_1\,e^{D_0/(RT)}}_{q_{\mathrm{elec}}}. }

(20)

where the thermal de Broglie wavelength is

\Lambda \equiv \sqrt{\frac{h^2}{2\pi m k_{\mathrm B}T}}.

(21)

Notes on parameters¶

$m$ : molecular mass.
$\sigma$ : rotational symmetry number — $\sigma=2$ for homonuclear diatomics (H $_2$ , I $_2$ ), $\sigma=1$ for heteronuclear (HI). See § 2.8.
$\Theta_{\mathrm{rot}} \equiv \hbar^2/(2 I k_{\mathrm B})$ : characteristic rotational temperature (from § 2.7).
$\Theta_{\mathrm{vib}} \equiv h\nu/k_{\mathrm B}$ : characteristic vibrational temperature (from § 2.6).
$g_1$ : ground-state electronic degeneracy. For the species in the HI reaction, all three ground states are $^1\Sigma^+$ singlets, so $g_1 = 1$ for each.
$D_0$ : bond dissociation energy referenced from the $v=0$ level (per-mole units when combined with $RT$ , per-molecule when combined with $k_{\mathrm B}T$ ).

7.2.7 $K$ for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ in terms of molecular parameters¶

Substituting Eq. (20) into $K = q_{HI}^2/(q_{H_2}\,q_{I_2})$ , and using the fact that volume cancels when $\Delta\nu=0$ (see Worked Example), yields

\boxed{ K = \left(\frac{m_{HI}^2}{m_{H_2}\,m_{I_2}}\right)^{3/2} \; \frac{\sigma_{H_2}\sigma_{I_2}}{\sigma_{HI}^2}\, \frac{\Theta_{\mathrm{rot}}^{H_2}\,\Theta_{\mathrm{rot}}^{I_2}}{\left(\Theta_{\mathrm{rot}}^{HI}\right)^2} \; \frac{\bigl(1-e^{-\Theta_{\mathrm{vib}}^{H_2}/T}\bigr)\bigl(1-e^{-\Theta_{\mathrm{vib}}^{I_2}/T}\bigr)}{\bigl(1-e^{-\Theta_{\mathrm{vib}}^{HI}/T}\bigr)^2} \; \exp\!\left(\frac{2D_0^{HI}-D_0^{H_2}-D_0^{I_2}}{RT}\right). }

(22)

With $\sigma_{H_2}=\sigma_{I_2}=2$ and $\sigma_{HI}=1$ , the symmetry-number prefactor is $\sigma_{H_2}\sigma_{I_2}/\sigma_{HI}^2 = 4$ ; vibrational zero-point factors have cancelled in pairs under the chosen energy zero, leaving only the $(1-e^{-\Theta_{\mathrm{vib}}/T})$ “hot-mode” factors.

Equation (22) makes the microscopic physics very explicit:

Translation contributes the $m^{3/2}$ mass dependence.
Rotation contributes the symmetry-number and $\Theta_{\mathrm{rot}}$ factors.
Vibration contributes the $(1-e^{-\Theta_{\mathrm{vib}}/T})$ factors that encode how many vibrational quanta are thermally accessible.
Bond strengths enter through the dissociation energies $D_0$ in the exponential; the numerator $2D_0^{HI}-D_0^{H_2}-D_0^{I_2}$ is essentially the negative of the reaction internal energy at 0 K.

7.2.8 Numerical evaluation: $K(700\,\mathrm{K})$ from spectroscopic data¶

To see the microscopic-to-macroscopic bridge at work, we evaluate Eq. (22) at $T = 700\ \mathrm{K}$ using standard spectroscopic constants (e.g., from Herzberg or the NIST WebBook):

species	$M$ (g/mol)	$\Theta_{\mathrm{rot}}$ (K)	$\Theta_{\mathrm{vib}}$ (K)	$D_0$ (kJ/mol)	$\sigma$	$g_1$
$\mathrm{H_2}$	2.016	87.6	6332	432.07	2	1
$\mathrm{I_2}$	253.808	0.0537	308	148.81	2	1
$\mathrm{HI}$	127.912	9.25	3266	294.67	1	1

import numpy as np

R = 8.314462618      # J mol^-1 K^-1

# Spectroscopic / thermochemical constants
mol = {
    "H2": dict(M=2.016e-3,   Theta_rot=87.6,   Theta_vib=6332.0, D0=432.07e3, sigma=2),
    "I2": dict(M=253.808e-3, Theta_rot=0.0537, Theta_vib=308.0,  D0=148.81e3, sigma=2),
    "HI": dict(M=127.912e-3, Theta_rot=9.25,   Theta_vib=3266.0, D0=294.67e3, sigma=1),
}
T = 700.0  # K

# Mass factor (translational)
mass_factor = (mol["HI"]["M"]**2 / (mol["H2"]["M"] * mol["I2"]["M"]))**1.5

# Rotational factor
rot_factor = (mol["H2"]["sigma"] * mol["I2"]["sigma"] / mol["HI"]["sigma"]**2) \
             * (mol["H2"]["Theta_rot"] * mol["I2"]["Theta_rot"]
                / mol["HI"]["Theta_rot"]**2)

# Vibrational factor: "hot-mode" (1 - exp(-Th_vib/T)) factors
def vfac(Tvib): return 1 - np.exp(-Tvib/T)
vib_factor = (vfac(mol["H2"]["Theta_vib"]) * vfac(mol["I2"]["Theta_vib"])
              / vfac(mol["HI"]["Theta_vib"])**2)

# Bond-energy factor: exp((2 D0_HI - D0_H2 - D0_I2)/RT)
dD0 = 2*mol["HI"]["D0"] - mol["H2"]["D0"] - mol["I2"]["D0"]  # J/mol
bond_factor = np.exp(dD0 / (R*T))

K = mass_factor * rot_factor * vib_factor * bond_factor

print(f"T = {T:.0f} K")
print(f"  Mass (translational) factor:       {mass_factor:10.3f}")
print(f"  Rotational factor:                 {rot_factor:10.4f}")
print(f"  Vibrational factor:                {vib_factor:10.4f}")
print(f"  Bond-energy factor exp(dD0/RT):    {bond_factor:10.4f}")
print(f"  2 D0(HI) - D0(H2) - D0(I2):        {dD0/1e3:+.2f} kJ/mol")
print()
print(f"  K(700 K) from molecular constants: {K:.1f}")
print(f"  Experimental K (Bodenstein, ~700 K): ~54")

T = 700 K
  Mass (translational) factor:          180.817
  Rotational factor:                     0.2199
  Vibrational factor:                    0.3627
  Bond-energy factor exp(dD0/RT):        4.2784
  2 D0(HI) - D0(H2) - D0(I2):        +8.46 kJ/mol

  K(700 K) from molecular constants: 61.7
  Experimental K (Bodenstein, ~700 K): ~54

Discussion. The predicted $K \approx 62$ compares to an experimental value near 54 at 700 K. Agreement at the ~15 % level is about what one should expect from the rigid-rotor/harmonic-oscillator approximation combined with uncertainties in $\Theta_{\mathrm{rot}}$ , $\Theta_{\mathrm{vib}}$ , and $D_0$ — corrections for anharmonicity, vibration–rotation coupling, and centrifugal distortion (none of which are in our model) would each shift $K$ by a few percent.

More importantly, the decomposition makes the physics transparent:

The translational mass factor ( $\approx 181$ ) is the largest single contribution. The ratio $m_{HI}^2/(m_{H_2}m_{I_2}) = (m_H+m_I)^2/(4 m_H m_I)$ is driven far from unity when the two atomic masses differ by a lot — for H and I, $m_I/m_H \approx 127$ , so $m_{HI}^2 \gg 4\,m_H m_I$ . If instead the atoms had similar masses (e.g., $\mathrm{D_2+X_2}\rightleftharpoons 2\mathrm{DX}$ with $m_D\sim m_X$ ), this factor would be close to 1.
The rotational factor ( $\approx 0.22$ ) is well below unity because I $_2$ is unusually heavy and has an extraordinarily small $\Theta_{\mathrm{rot}} \approx 0.05$ K; the corresponding $q_{\mathrm{rot}}(\mathrm{I_2})$ is very large, which favors keeping mass in I $_2$ rather than redistributing it into HI. The factor-of-4 symmetry-number prefactor (from $\sigma_{H_2}\sigma_{I_2}/\sigma_{HI}^2$ ) nudges equilibrium toward products but doesn’t overcome this.
The vibrational factor ( $\approx 0.36$ ) is also below unity. HI has a stiff H–I stretch that is still nearly frozen out at 700 K (numerator small for HI), while I $_2$ ’s low-frequency mode is partially excited (numerator moderate for I $_2$ ), and these combine unfavorably for products.
The bond-energy factor ( $\approx 4.3$ ) contains the “chemistry”: $2D_0^{HI} - D_0^{H_2} - D_0^{I_2} \approx +8.5\,\mathrm{kJ/mol}$ says the two HI bonds together are slightly stronger than one H–H plus one I–I bond, a modest exothermicity that at 700 K gives just over a factor of 4 in $K$ .

The lesson: every digit of $K$ can be traced back to a specific molecular property, and the relative sizes of the four factors tell us which part of molecular structure matters most for this particular equilibrium. A student who computed $\Delta_r G^\circ$ from thermochemical tables (§ 7.1.6) and a student who computed $K$ from spectroscopic constants (§ 7.2.8) would arrive at the same number by very different routes — and each route teaches something different about why the equilibrium sits where it does.

Worked Example¶

Why volume cancels when $\Delta\nu=0$ : $\mathrm{H_2+I_2\rightleftharpoons 2HI}$ ¶

Writing each single-molecule partition function as $q_i = q_{\mathrm{trans},i}\,q_{\mathrm{int},i}$ with $q_{\mathrm{trans},i}=V/\Lambda_i^3$ (and $q_{\mathrm{int},i}$ collecting the rotational, vibrational, and electronic factors, all of which are $V$ -independent):

K = \frac{q_{HI}^2}{q_{H_2}\,q_{I_2}} = \frac{(V/\Lambda_{HI}^3)^2\,q_{\mathrm{int},HI}^2}{(V/\Lambda_{H_2}^3)(V/\Lambda_{I_2}^3)\,q_{\mathrm{int},H_2}\,q_{\mathrm{int},I_2}}.

(23)

Because $\Delta\nu = 2-1-1 = 0$ , the volume factors cancel:

\frac{V^2}{V\cdot V}=1,

(24)

leaving

K=\left(\frac{\Lambda_{H_2}^3\Lambda_{I_2}^3}{\Lambda_{HI}^6}\right)\, \frac{q_{\mathrm{int},HI}^2}{q_{\mathrm{int},H_2}\,q_{\mathrm{int},I_2}}.

(25)

Since $\Lambda\propto m^{-1/2}$ , the translational contribution collapses to the mass factor $\left(m_{HI}^2/(m_{H_2}m_{I_2})\right)^{3/2}$ seen in Eq. (22).

Result. For reactions with $\Delta\nu=0$ , $K$ is independent of volume (and pressure) in the ideal-gas model; microscopic physics enters through masses and internal partition functions only. For reactions with $\Delta\nu\ne 0$ , the uncancelled factors of $V$ reappear as the $(k_{\mathrm B}T/P^\circ)^{\Delta\nu}$ conversion in Eq. (15).

Concept Checks¶

Why is the Helmholtz free energy the relevant potential for equilibrium at constant $T,V$ , and why does the resulting condition $\sum_i\nu_i\mu_i=0$ match the condition derived in § 7.1 at constant $T,P$ ?
Where does Stirling’s approximation enter the derivation of $\mu_i = -RT\ln(q_i/N_i)$ , and what physical limit makes it reasonable?
In Eq. (20), how would the vibrational factor change if we chose the energy zero at the bottom of the potential well rather than at $v=0$ ? Show that $K$ is unchanged by this rechoice (as it must be).
Vibrational contributions to $K$ often matter most at high temperatures. Why? Reason from the temperature dependence of $(1-e^{-\Theta_{\mathrm{vib}}/T})$ .
Suppose all three species in a $\Delta\nu=0$ reaction had identical masses, moments of inertia, vibrational frequencies, and $g_1$ values, but different $D_0$ values. What would $K$ reduce to, and what does this limit teach you about when “bond counting” arguments are quantitatively trustworthy?

Key Takeaways¶

Microscopic partition functions provide a route to equilibrium constants by linking $A = -k_{\mathrm B}T\ln Q$ (§ 4.3) to chemical potentials via $\mu_i = (\partial A/\partial N_i)_{T,V}$ .
For ideal-gas mixtures, $\mu_i = -RT\ln(q_i/N_i)$ (per-mole), and the equilibrium condition $\sum_i\nu_i\mu_i=0$ delivers $K$ as a ratio of single-molecule partition functions.
Translational, rotational, vibrational, and electronic factors contribute in separable ways under the common rigid-rotor/harmonic-oscillator approximations of Chapter 2.
For $\Delta\nu=0$ gas reactions, volume (and hence pressure) dependence cancels in the ideal-gas equilibrium constant; for $\Delta\nu\ne 0$ , the $(k_{\mathrm B}T/P^\circ)^{\Delta\nu}$ factor converts between $K_c$ and $K_p$ .
For the $\mathrm{H_2+I_2\rightleftharpoons 2HI}$ benchmark, the four factors in Eq. (22) predict $K(700\,\mathrm{K})\approx 62$ , in ~15 % agreement with the experimental value of ~54 — a concrete illustration of how molecular structure alone determines the position of equilibrium.