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Reactions on surfaces

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Reactions involving the adsorption of at least a reactant onto a surface

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Simple decomposition

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If a reaction occurs through these steps:

A + S ⇌ AS → Products

where A is the reactant and S is an adsorption site on the surface and the respectiverate constants for the adsorption, desorption and reaction arek₁,k₋₁ andk₂, then the global reaction rate is:

r=k_{2}C_{\mathrm {AS} }=k_{2}\theta C_{\mathrm {S} }

where:

r is the rate, mol·m⁻²·s⁻¹
$C_{A}$ is the concentration ofadsorbate,mol·m⁻³
$C_{\mathrm {AS} }$ is the surface concentration of occupied sites,mol·m⁻²
$C_{\mathrm {S} }$ is the concentration of all sites (occupied or not),mol·m⁻²
$\theta$ is thesurface coverage, (i.e. $C_{AS}/C_{S}$ ) defined as the fraction of sites which are occupied, which is dimensionless
$t {\displaystyle t}$ is time,s
$k_{2}$ is the rate constant for the surface reaction,s⁻¹.
$k_{1}$ is the rate constant for surface adsorption,m³·mol⁻¹·s⁻¹
$k_{-1}$ is the rate constant for surface desorption,s⁻¹

$C_{\mathrm {S} }$ is highly related to the total surface area of the adsorbent: the greater the surface area, the more sites and the faster the reaction. This is the reason why heterogeneous catalysts are usually chosen to have great surface areas (in the order of a hundredm²/gram)

If we apply thesteady state approximation to AS, then:

{\frac {dC_{\mathrm {AS} }}{dt}}=0=k_{1}C_{\mathrm {A} }C_{\mathrm {S} }(1-\theta )-k_{2}\theta C_{\mathrm {S} }-k_{-1}\theta C_{\mathrm {S} }

\theta ={\frac {k_{1}C_{\mathrm {A} }}{k_{1}C_{\mathrm {A} }+k_{-1}+k_{2}}}

and

r={\frac {k_{1}k_{2}C_{\mathrm {A} }C_{\mathrm {S} }}{k_{1}C_{\mathrm {A} }+k_{-1}+k_{2}}}.

The result is equivalent to theMichaelis–Menten kinetics of reactions catalyzed at a site on anenzyme. The rate equation is complex, and the reactionorder is not clear. In experimental work, usually two extreme cases are looked for in order to prove the mechanism. In them, therate-determining step can be:

Limiting step: adsorption/desorption

k_{2}\gg \ k_{1}C_{\mathrm {A} },k_{-1},{\text{ so }}r\approx k_{1}C_{\mathrm {A} }C_{\mathrm {S} }.

The order respect to A is 1. Examples of this mechanism areN₂O on gold andHI onplatinum

Limiting step: reaction of adsorbed species

k_{2}\ll \ k_{1}C_{\mathrm {A} },k_{-1}{\text{ so }}\theta ={\frac {k_{1}C_{\mathrm {A} }}{k_{1}C_{\mathrm {A} }+k_{-1}}}={\frac {K_{1}C_{\mathrm {A} }}{K_{1}C_{\mathrm {A} }+1}}

The last expression is theLangmuir isotherm for the surface coverage. The adsorption equilibrium constant $K_{1}={\frac {k_{1}}{k_{-1}}}$ , and the numerator and denominator have each been divided by $k_{-1}$ . The overall reaction rate becomes $r={\frac {K_{1}k_{2}C_{\mathrm {A} }C_{\mathrm {S} }}{K_{1}C_{\mathrm {A} }+1}}$ .

Depending on the concentration of the reactant the rate changes:

Low concentrations, then $r=K_{1}k_{2}C_{\mathrm {A} }C_{\mathrm {S} }$ , that is to say a first order reaction in component A.
High concentration, then $r=k_{2}C_{\mathrm {S} }$ . It is a zeroth order reaction in component A.

Bimolecular reaction

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Langmuir–Hinshelwood mechanism

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In this mechanism, suggested byIrving Langmuir in 1921 and further developed byCyril Hinshelwood in 1926, two molecules adsorb on neighboring sites and the adsorbed molecules undergo a bimolecular reaction:^[3]

A + S ⇌ AS

B + S ⇌ BS

AS + BS → Products

The rate constants are $k_{1}$ and $k_{-1}$ for adsorption and desorption of A respectively, $k_{2}$ and $k_{-2}$ for adsorption and desorption of B, and $k {\displaystyle k}$ for the reaction generating the final products. The rate law is: $r=k\theta _{\mathrm {A} }\theta _{\mathrm {B} }C_{\mathrm {S} }^{2}$

Proceeding as before we get $\theta _{\mathrm {A} }={\frac {k_{1}C_{\mathrm {A} }\theta _{E}}{k_{-1}+kC_{\mathrm {S} }\theta _{\mathrm {B} }}}$ , where $\theta _{E}$ is the fraction of empty sites, so $\theta _{\mathrm {A} }+\theta _{\mathrm {B} }+\theta _{E}=1$ . Let us assume now that the rate limiting step is the reaction of the adsorbed molecules, which is easily understood: the probability of two adsorbed molecules colliding is low.Then $\theta _{\mathrm {A} }=K_{1}C_{\mathrm {A} }\theta _{E}$ , with $K_{i}=k_{i}/k_{-i}$ , which is nothing but Langmuir isotherm for two adsorbed gases, with adsorption constants $K_{1}$ and $K_{2}$ .Calculating $\theta _{E}$ from $\theta _{\mathrm {A} }$ and $\theta _{\mathrm {B} }$ we finally get

r=kC_{\mathrm {S} }^{2}{\frac {K_{1}K_{2}C_{\mathrm {A} }C_{\mathrm {B} }}{(1+K_{1}C_{\mathrm {A} }+K_{2}C_{\mathrm {B} })^{2}}}

The rate law is complex and there is no clear order with respect to either reactant, but we can consider different values of the constants, for which it is easy to measure integer orders:

Both molecules have low adsorption

That means that $1\gg K_{1}C_{\mathrm {A} },K_{2}C_{\mathrm {B} }$ , so $r=kC_{\mathrm {S} }^{2}K_{1}K_{2}C_{\mathrm {A} }C_{\mathrm {B} }$ . The order is one with respect to each reactant, and the overall order is two.

One molecule has very low adsorption

In this case $K_{1}C_{\mathrm {A} },1\gg K_{2}C_{\mathrm {B} }$ , so $r=kC_{\mathrm {S} }^{2}{\frac {K_{1}K_{2}C_{\mathrm {A} }C_{\mathrm {B} }}{(1+K_{1}C_{\mathrm {A} })^{2}}}$ . The reaction order is 1 with respect to B. There are two extreme possibilities for the order with respect to A:

At low concentrations of A, $r=kC_{\mathrm {S} }^{2}K_{1}K_{2}C_{\mathrm {A} }C_{\mathrm {B} }$ , and the order is one with respect to A.
At high concentrations, $r=kC_{\mathrm {S} }^{2}{\frac {K_{2}C_{\mathrm {B} }}{K_{1}C_{\mathrm {A} }}}$ . The order is minus one with respect to A. The higher the concentration of A, the slower the reaction goes, in this case we say that A inhibits the reaction.

One molecule has very high adsorption

One of the reactants has very high adsorption and the other one doesn't adsorb strongly.

$K_{1}C_{\mathrm {A} }\gg 1,K_{2}C_{\mathrm {B} }$ , so $r=kC_{\mathrm {S} }^{2}{\frac {K_{2}C_{\mathrm {B} }}{K_{1}C_{\mathrm {A} }}}$ . The reaction order is 1 with respect to B and −1 with respect to A. Reactant A inhibits the reaction at all concentrations.

The following reactions follow a Langmuir–Hinshelwood mechanism:^[4]

2CO +O₂ → 2CO₂ on aplatinum catalyst.
CO +2H₂ →CH₃OH on aZnO catalyst.
C₂H₄ + H₂ →C₂H₆ on acopper catalyst.
N₂O + H₂ →N₂ + H₂O on a platinum catalyst.
C₂H₄ +⁠1/2⁠ O₂ →CH₃CHO on apalladium catalyst.
CO + OH → CO₂ + H⁺ + e⁻ on a platinum catalyst.

Langmuir–Rideal mechanism

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In this mechanism, proposed in 1922 byIrving Langmuir and later expanded upon byEric Rideal, only one of the molecules adsorbs and the other one reacts with it directly from the gas phase, without adsorbing ("nonthermal surface reaction"):

A(g) + S(s) ⇌ AS(s)

AS(s) + B(g) → Products

Constants are $k_{1},k_{-1}$ and $k {\displaystyle k}$ and rate equation is $r=kC_{\mathrm {S} }\theta _{\mathrm {A} }C_{\mathrm {B} }$ . Applying steady state approximation to AS and proceeding as before (considering the reaction the limiting step once more) we get $r=kC_{\mathrm {S} }C_{\mathrm {B} }{\frac {K_{1}C_{\mathrm {A} }}{K_{1}C_{\mathrm {A} }+1}}$ . The order is one with respect to B. There are two possibilities, depending on the concentration of reactant A:

At low concentrations of A, $r=kC_{\mathrm {S} }K_{1}C_{\mathrm {A} }C_{\mathrm {B} }$ , and the order is one with respect to A.

At high concentrations of A, $r=kC_{\mathrm {S} }C_{\mathrm {B} }$ , and the order is zero with respect to A.

The following reactions follow an Langmuir–Rideal mechanism:^[4]

C₂H₄ +⁠1/2⁠ O₂ (adsorbed) →(CH₂CH₂)O Thedissociative adsorption of oxygen is also possible, which leads to secondary productscarbon dioxide andwater.
CO₂ + H₂ (ads.) → H₂O + CO
2NH₃ +⁠1+1/2⁠ O₂ (ads.) → N₂ + 3H₂O on a platinum catalyst
C₂H₂ + H₂ (ads.) → C₂H₄ onnickel oriron catalysts

The Langmuir-Rideal mechanism is often, incorrectly, attributed toDan Eley as the Eley-Rideal mechanism.^[5] The actual Eley-Rideal mechanism, studied in the thesis ofDan Eley and proposed byEric Rideal in 1939, was the reaction between achemisorbed and aphysisorbed molecule.^[6] As opposed to the Langmuir-Rideal mechanism, in this mechanism the physisorbed molecule is in thermal equilibrium with the surface.