Understanding the Temperature Dependence of TMA → O-Al(OH)₂* ALD

Developed by Courtney Palmeri (BS '26) · University of Rochester

1. Molecular-Level Visualization of One ALD Cycle

Layers: 0 (bare)

Water vapor (H₂O) flows in from the left, adsorbs onto bare surface sites, and dissociates to form OH* groups. Adsorption rate scales with |ΔH_ads|.

Surface Site (*)

OH*

O-Al(CH₃)₂*

O-Al(OH)₂*

CH₄ / H₂O / TMA (gas)

2. Rate vs Temperature

3. O-Al(CH₃)₂* Coverage θ_A vs Temperature (ALD effectiveness)

4. Energy Diagram (per reaction pathway)

5. Heat Removal Q vs Temperature (rate of energy released)

6. Reaction Mechanism (simplified: θ_* = 0 after Step 1)

(1) $\text{OH*} + \text{Al(CH}_3)_3(g) \;\rightleftharpoons\; \text{O-Al(CH}_3)_2\text{*} + \text{CH}_4(g)$ (reversible, equilibrium $K_1$)

(2) $\text{O-Al(CH}_3)_2\text{*} + 2\text{H}_2\text{O}(g) \;\rightarrow\; \text{O-Al(OH)}_2\text{*} + 2\text{CH}_4(g)$ (irreversible, rate-determining)

7. Governing Equations

Surface Site Balance $\theta_{OH} + \theta_A = 1$ Step 1 Equilibrium ($K_1$) $K_1 = \dfrac{\theta_A\, P_{CH_4}}{\theta_{OH}\, P_{TMA}}$

$\theta_A = \dfrac{K_1\, P_{TMA}}{P_{CH_4} + K_1\, P_{TMA}}$

Rate Law

$-r_2 = k_2(T)\,P_{H_2O}^{2}\;\dfrac{K_1\,P_{TMA}}{P_{CH_4} + K_1\,P_{TMA}}$

Rate-determining step only

Temperature Dependence

$k(T) = k_{ref}\,\exp\!\left(-\dfrac{E_a}{R}\left(\dfrac{1}{T}-\dfrac{1}{T_{ref}}\right)\right)$

$K(T) = K_{ref}\,\exp\!\left(-\dfrac{\Delta H}{R}\left(\dfrac{1}{T}-\dfrac{1}{T_{ref}}\right)\right)$

$R = 8.314\;$J/mol·K, $T_{ref}=300\;$K

Coverage

$\theta_A \to 1$ ⇒ TMA-saturated surface

$\theta_A \to 0$ ⇒ OH-terminated

$\theta_A$ peaks where $K_1\,P_{TMA} \gg P_{CH_4}$

Energy Balance (isothermal)

$Q = -V\,\Delta H_3\,(-r_2)$

Heat capacities (25 °C):

$C_{p,TMA} = 155.6$ J/mol·K

$C_{p,H_2O} = 4.184$ J/g·°C

$C_{p,subs} = 2.32$ J/kg·K

$\Delta H<0\;\Rightarrow$ Exothermic

$Q>0\;\Rightarrow$ Heat Removed