MICROKINETIC MODEL OF NH₃ DISSOCIATION AND N₂ AND H₂ FROMATION

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Initial conditions

In this model, the initial temperature T₀ start at 50K, and a heating rate Beta = 3 K/s.

Ammonia adsorption and desorption step

• NH₃ (gas) + * → NH₃* (R1), k₁ $$ r_1=k_1 P_{NH3}\theta $$

NH₃ dissociation is done in 3 elementary steps

• NH₃* + * ⇄ NH₂* + H* (R2), k_f2, k_b2 $$ r_2 = k_{f2}\theta_{NH3}\theta-k_{b2}\theta_{NH2}\theta_{H} $$

• NH₂* + * ⇄ NH* + H* (R3), k_f3, k_b3 $$ r_3 = k_{f3}\theta_{NH2}\theta-k_{b3}\theta_{NH}\theta_{H} $$

• NH* + * ⇄ N* + H* (R4), k_f4, k_b4 $$ r_4 = k_{f4}\theta_{NH}\theta-k_{b4}\theta_{N}\theta_{H} $$

N₂ formation is done in 3 elementary steps

• 2 N* ⇄ N-N****** (R5), k_f5, k_b5 $$ r_5=k_{f5}\theta_N^2 - k_{b5}\theta_{N-N} $$

• N-N****** → *N₂ + * (R6), k₆ $$ r_6= k_6\theta_{N-N} $$

• N₂* → N₂ (gas) + * (R7), k₇ $$ r_7= k_7 \theta_{N2} $$

H₂ formation is done in one elementary step

• 2 H* → H₂ (gas) + 2* (R8), k₈ $$ r_8= k_8 \theta_H^2 $$

Rate constant equations

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Surface reactions

$$ k_{sur} =\frac{k_BT}{h}\ exp(\frac{-E_a}{k_BT}) $$

Desorption

$$ k_{des} = \frac{k_BT^3}{h^3} \frac{A(2\pi m k_B)}{\sigma \theta {rot}} exp(\frac{-E{des}}{k_B T}) $$

Rate equations of the different species

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$$ \frac{dP_{NH3}}{dt} = -r_1 $$

$$ \frac{d\theta_{NH3}}{dt} = r_1-r_2 $$

$$ \frac{d\theta_{NH2}}{dt} = r_2-r_3 $$

$$ \frac{d\theta_{NH}}{dt} = r_3-r_4 $$

$$ \frac{d\theta_N}{dt}= r_4 - 2r_5 $$

$$ \frac{d\theta_{N-N}}{dt}= r_5-r_6 $$

$$ \frac{d\theta_{N2}}{dt}= r_6-r_7 $$

$$ \frac{d\theta_{H}}{dt}= r_2+r_3+r_4 - 2r_8 $$

$$ \frac{d\theta}{dt}= r_6+r_7+2r_8-r_1-r_2-r_3-r_4 $$

$$ \frac{dP_{N2}}{dt}= r_5 $$

$$ \frac{dP_{H2}}{dt}= r_8 $$

$$ \frac{dT}{dt}= \beta $$

Steady State Approximation

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To simplify our current model, steady-state approximation of the following adsorbed species was used: NH_2ads, N-N_ads, N_2ads, and H_ads

$$ \frac{d\theta_{NH2}}{dt} =0 $$

$$ \frac{d\theta_{N-N}}{dt} =0 $$

$$ \frac{d\theta_{N2}}{dt} = 0 $$

$$ \frac{d\theta_{H}}{dt} =0 $$

Thus: $$ r_3 = r_2 $$

$$ r_6 = r_5 $$

$$ r_7 = r_5 $$

$$ r_8 = r_2 + 1/2 r_4 $$

which gives: $$ \frac{dP_{NH3}}{dt} = -r_1 $$

$$ \frac{d\theta_{NH3}}{dt} = r_1-r_2 $$

$$ \frac{d\theta_{NH2}}{dt} = 0 $$

$$ \frac{d\theta_{NH}}{dt} = r_3-r_4 $$

$$ \frac{d\theta_N}{dt}= r_4 - 2r_5 $$

$$ \frac{d\theta_{N-N}}{dt}= 0 $$

$$ \frac{d\theta_{N2}}{dt}= r_6-r_7 $$

$$ \frac{d\theta_{H}}{dt}= 0 $$

$$ \frac{d\theta}{dt}= 2r_5-r_1 $$

$$ \frac{dP_{N2}}{dt}= r_5 $$

$$ \frac{dP_{H2}}{dt}= r_2-1/2r_4 $$

$$ \frac{dT}{dt}= \beta $$

Solve the ODE

Because our model is stiff, the ODEs cannot be solved using an explicit method. Consequently, BDF method (Backward-Differentiation Formulas) was used instead of Runge Kutta. The former however requires a Jacobian matrix of the right-hand side of the system with respect to y. Thus, a 12 x 12 Jacobian matrix has to be define prior to solving the ODEs.