Lattice vibrations contribute a phonon heat capacity. At temperatures well below the Debye temperature, its leading contribution is proportional to $T^3$, whereas the electronic part is proportional to $T$:The lattice term can therefore become important rapidly as temperature rises, even if the electronic term dominates sufficiently close to zero. At higher temperature the lattice contribution approaches the Dulong-Petit law value $3R$ per mole of atoms within the harmonic approximation. This is not a claim that the low-temperature $T^3$ law holds at all temperatures.

The Fermi temperature isInserting it into the question's quoted estimate givesAs explained in (d), that count is per mole of conduction electrons. Using the stipulated six electrons per tungsten atom gives $1.8025\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ per mole of tungsten in the same rough model. The accurate free-electron Sommerfeld expansion coefficient would instead give $4.4474\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ per mole of tungsten. These are different approximations and counting conventions, not three interchangeable exact answers.

For $T\ll T_F$, only electrons within an energy window of order $kT$ around the Fermi energy can change occupation appreciably, because of the Pauli exclusion principle. The active fraction is of order $kT/E_F=T/T_F$. Giving each of these electrons energy of order $kT$ estimates an excitation energy $U-U_0\sim N_{e}k{T}^2/T_F$, and thereforePer mole of conduction electrons, $N_{e}k=R$, which is the quoted $2RT/T_F$. Per mole of metal atoms with $z$ model conduction electrons each, the conduction-electron count in the molar electronic heat capacity instead gives $2zRT/T_F$. The numerical factor is only approximate: the Sommerfeld expansion gives $C_V=({\pi }^2/2)N_{e}kT/T_F$ for a three-dimensional free-electron gas.

The Fermi sphere volume in reciprocal space isThe primitive real-space cell has volume $a^3/2$, so the first Brillouin zone has volumeEach band accommodates two electrons per primitive cell when spin is counted. Six electrons therefore require three such zone-volumes of occupied reciprocal-space states in the free-electron counting picture. The extended-zone sphere is not confined to the first zone; folding it produces multiple bands. It does not imply that the actual tungsten Fermi surface is spherical.

There are twelve model conduction electrons in the conventional unit cell. Hence $n=12/a^3=3.78495\times 10^{29}{\mathrm{m}}^{-3}$. For a spin-twofold free electron gas,Numerically,The six-electron count is the stipulated model; treating transition-metal d states as completely free is an approximation.

In a body-centred cubic lattice, touching atoms lie along a body diagonal. Its length is $\sqrt{3}a$ and spans four radii, so

Identify the transition-state separation with contact, $R=r_A+r_B$, take distinct atoms so $\sigma =1$, and let $E_A=N_A\Delta {\epsilon }_0^{‡}$. Then $\pi R^2=\rho$ and $\Delta {\epsilon }_0^{‡}/kT=E_A/RT$, yieldingThe velocity factor is the mean relative speed from the Maxwell-Boltzmann distribution. Thus this simple transition state theory model reproduces the stated collision theory expression. Molecular orientation constraints, quantum transmission and reaction-surface recrossing can invalidate the equality outside these simplifying assumptions.

As a first estimate, neglect the small mass change and changes in geometry and vibrational populations. Removing an electron creates an unpaired spin, giving at least the spin-only electronic entropy increaseFor a more specific estimate, the actual cation ground term is $X{}^2{\Pi }_g$, rather than an orbitally nondegenerate doublet. Ignoring spin-orbit coupling, its two orbital components and two spin components give electronic degeneracy four andThe ground-state assignment is confirmed by state-resolved photoionization measurements.^[ref] Spin-orbit splitting and changed vibrational/rotational constants refine this estimate; the question's supplied partition functions alone do not determine those corrections. The $R\log 2$ answer is a spin-only estimate, while $R\log 4$ additionally retains the orbital degeneracy.