Q2(d). Surface-catalysed decomposition: zero order at high pressure
Show that a surface catalysed decomposition reaction follows zero order kinetics with respect to reactant when its pressure is high enough.
Answer in brief
Only adsorbed molecules react, so the rate is proportional to the fraction of surface covered, given by the Langmuir isotherm: rate =kθ=(kKp)/(1+Kp). At high pressure Kp≫1, θ→1 and rate =k, independent of p: zero order.
Key points
- The reactant A is adsorbed quickly and reversibly on a vacant site S of the catalyst; the adsorbed molecule decomposes slowly; the products are weakly held and leave at once: 4pt] A(g) + S ⇌A-S (fast, equilibrium) A-S…
- a monolayer; all sites equivalent, one molecule per site; no interaction between adsorbed molecules; adsorption and desorption in dynamic equilibrium.
- rate of adsorption =kₐ p (1-θ), rate of desorption =k_d θ. At equilibrium, with K=kₐ/k_d: θ=(Kp)/(1+Kp) (Langmuir isotherm)
- Only adsorbed molecules decompose, so the rate is proportional to θ: v=-(dp)/(dt)=kθ=(kKp)/(1+Kp)
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Mechanism and rate law
- The reactant A is adsorbed quickly and reversibly on a vacant site S of the catalyst; the adsorbed molecule decomposes slowly; the products are weakly held and leave at once: 4pt] A(g) + S ⇌A-S (fast, equilibrium) A-S →products + S (slow)
- Langmuir assumptions: a monolayer; all sites equivalent, one molecule per site; no interaction between adsorbed molecules; adsorption and desorption in dynamic equilibrium.
- With θ the fraction of sites covered: rate of adsorption =kₐ p (1-θ), rate of desorption =k_d θ. At equilibrium, with K=kₐ/k_d: θ=(Kp)/(1+Kp) (Langmuir isotherm)
- Only adsorbed molecules decompose, so the rate is proportional to θ: v=-(dp)/(dt)=kθ=(kKp)/(1+Kp)
The two limits
- High pressure, Kp≫1: 1+Kp≈ Kp, so θ→1 and v=k (zero order: rate independent of p) Every site is occupied (surface saturated); adding more gas cannot add molecules to the surface, so the rate is fixed by the number of sites and k. Integrated: p=p₀-kt, a straight line; t_(1/2)=p₀/2k grows with p₀.
- Low pressure, Kp≪1: θ≈ Kp and v=kKp: first order; the surface is nearly bare and the number of adsorbed molecules is proportional to p.
- In between: the order n=(dln v)/(dln p)=(1)/(1+Kp) falls smoothly from 1 to 0; at Kp=1 it is ½. A fit v=k'pⁿ with 0<n<1 (fractional order) corresponds to a Freundlich isotherm.
- Strongly adsorbed gases (large K) reach the zero-order limit at low pressures: NH₃ on tungsten or molybdenum and HI on gold decompose at a rate independent of pressure. PH₃ on tungsten is first order at low pressure and zero order at high pressure.
- Linear test: 1/v=1/k+1/kK·1/p; a plot of 1/v against 1/p is a straight line, intercept 1/k, slope 1/kK.
Temperature: in the zero-order region the measured rate constant is the true k, so the Arrhenius plot gives the true activation energy of the surface step; at low pressure it gives kK, and the apparent activation energy is lowered by the (negative) enthalpy of adsorption.
Application. NH₃ decomposition on a hot tungsten wire at high pressure is the classic zero-order reaction; in industry the same saturation means that raising the reactant pressure above a point gives no gain in rate, whereas raising the catalyst area does (finely divided metals, supported catalysts).
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All parts of Question 2
- 2(a) Heat absorbed in an isothermal expansion against constant pressure10 marks
- 2(b) Heat engine between 500 K and 300 K: is 450 J from 1000 J possible?10 marks
- 2(c) Opposing first-order reactions A ⇌ B with k₁/k₂=215 marks
- 2(d) Surface-catalysed decomposition: zero order at high pressure15 marks