Write the Morse potential relative to its minimum as

V (a + x) = D [e^{- 2 x / ξ} - 2 e^{- x / ξ}]

. Expanding one order beyond the harmonic term gives

V (a + x) = - D + \frac{D}{ξ ^{2}} x^{2} - \frac{D}{ξ ^{3}} x^{3} + O (x^{4}) = - D + \frac{κ}{2} x^{2} - \frac{κ}{2 ξ} x^{3} + O (x^{4}) .

The negative cubic coefficient means that stretching is softer than compression. Thermal motion therefore has positive mean displacement even though the most probable minimum is at

x = 0

⟨ x^{2} ⟩_{0} = k_{B} T / κ

. Vanishing mean restoring force, to first anharmonic order, requires

0 = ⟨ V^{'} (a + x)⟩ = κ ⟨ x ⟩ - \frac{3 κ}{2 ξ} ⟨ x^{2} ⟩ + \dots, ⟨ x ⟩ ≃ \frac{3 k _{B} T}{2 κ ξ} .

For a bond free to expand under negligible applied tension,

L = ⟨ r ⟩ ≃ a + ⟨ x ⟩

α \sim \frac{k _{B}}{aκ ξ} ≃ 3.9 \times 1 0^{- 5} K^{- 1} .

Keeping the leading perturbative numerical factor gives

α ≃ 3 k_{B} / (2 aκ ξ) = 5.9 \times 1 0^{- 5} K^{- 1}

. The question only requires the scale, not that factor. The thermal expansion is positive; a purely harmonic approximation would give zero mean displacement and zero thermal expansion. The local Taylor expansion describes a bound bond and is not a globally stable cubic potential.