Question

The reverse water-gas shift (RWGS) reaction is an equimolar reaction between CO2 and H2 to form CO and H2O. Assume CO2 associatively adsorbs to the surface, while H2 dissociatively adsorbs. These adsorption steps are followed by reversible formation of formate (COOH*) and slow dissociation of formate into gaseous CO and adsorbed OH. The adsorbed OH is then removed as gaseous H2O via a hydrogenation step.
a) Using the details of the mechanism listed above, write out the elementary steps for the RWGS reaction.
b) Derive a rate law for the RWGS reaction consistent with the above assumptions and mechanism from (i).
c) Under what conditions is the RWGS reaction first order in CO2?

Answer 1

Answer:

a) Check explanation for this

b)Rate law is  $Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]$

c) The rate does not depend on the concentration of CO₂

Explanation:

a) Elementary steps for the RWGS reaction:

• Dissociative adsorption of the H₂ Molecule

                 $H_{2} \xrightarrow{\text{k1}} H + H$   (Fast process)

• Reversible Reaction between CO₂ and H

                $\[ CO_{2} + H\mathrel{\mathop{\rightleftarrows}^{\mathrm{k2}}_{\mathrm{k3}}}COOH \]$ (Fast Process)

• Slow dissociation of COOH into gaseous CO and absorbed OH

                $COOH \xrightarrow{\text{k1}} CO + OH$ (Slow process)

• Fast hydrogenation of the OH to form H₂O

                   $OH + H \xrightarrow{\text{k5}} H_{2} O$ (Fast process)

b) Derivation of the rate law

We need to determine the rate law for H, OH and COOH because these are the intermediates for this reaction.

The steady state approximation is applied to a consecutive reaction with a slow first step and a fast second step (k1≪k2). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product.

Rate of consumption = Rate of production

For COOH:

Using steady state approximation

$\frac{d[COOH]}{dt} = 0$

$k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]$

$[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} }$

For H:

$\frac{d[H]}{dt} = 0$

$k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]$

$[H]= \frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]}$

For OH:

$\frac{d[OH]}{dt} = 0$

$k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}$

The rate of the overall reaction is determined by the slowest step of the reaction. The slowest process is the dissociation of COOH

Therefore the overall rate of reaction is:

$Rate = k_{4} [COOH]$

$Rate = k_{4} \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} }\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]} }{k_{3}k_{4}}\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}\frac{k_{4}COOH }{k_{5}H } +k_{2}[CO_{2}]} }{k_{3}k_{4}}$

Simplifying the equation above, the rate law becomes

$Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]$

c) It is obvious from the rate law written above that the rate of the RWBG reaction does not depend on the concentration of CO₂