Answer:
a) Check explanation for this
b)Rate law is ![Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]](https://tex.z-dn.net/?f=Rate%20%3D%20%5Cfrac%7Bk_%7B1%7Dk_%7B4%7D%20%20%7D%7Bk_%7B3%7D%2B%202k_%7B4%7D%20%20%7D%20%5BH_%7B2%7D%20%5D)
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
(Fast process)
- Reversible Reaction between CO₂ and H
(Fast Process)
- Slow dissociation of COOH into gaseous CO and absorbed OH
(Slow process)
- Fast hydrogenation of the OH to form H₂O
(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](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BCOOH%5D%7D%7Bdt%7D%20%3D%200)
![k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]\\](https://tex.z-dn.net/?f=k_%7B2%7D%20%5BCO_%7B2%7D%20%5D%5BH%5D%20%3D%20k_%7B3%7D%20%5BCOOH%5D%20k_%7B4%7D%20%5BCOOH%5D%5C%5C)
![[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} } \\](https://tex.z-dn.net/?f=%5BCOOH%5D%20%3D%20%5Cfrac%7Bk_%7B2%7D%20%5BCO_%7B2%7D%20%5D%5BH%5D%7D%7Bk_%7B3%7Dk_%7B4%7D%20%20%7D%20%5C%5C)
For H:
![\frac{d[H]}{dt} = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D%20%3D%200)
![k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]](https://tex.z-dn.net/?f=k_%7B1%7D%5BH_%7B2%7D%5D%20%3D%20k_%7B2%7D%5BCO_%7B2%7D%20%5BH%5D%2Bk_%7B5%7D%20%5B%20OH%5D%5BH%5D)
![[H]= \frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]}\\](https://tex.z-dn.net/?f=%5BH%5D%3D%20%5Cfrac%7Bk_%7B1%7D%5BH_%7B2%7D%5D%20%20%7D%7Bk_%7B5%7D%5BOH%5D%20%2Bk_%7B2%7D%5BCO_%7B2%7D%5D%7D%5C%5C)
For OH:
![\frac{d[OH]}{dt} = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BOH%5D%7D%7Bdt%7D%20%3D%200)
![k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}\\](https://tex.z-dn.net/?f=k_%7B4%7D%20%5BCOOH%5D%20%3D%20k_%7B5%7D%20%5BOH%5D%5BH%5D%5C%5C%5Ck%5BOH%5D%20%3D%20%5Cfrac%7Bk_%7B4%7D%20%5BCOOH%5D%7D%7Bk_%7B5%7D%20H%7D%5C%5C)
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]\\](https://tex.z-dn.net/?f=Rate%20%3D%20k_%7B4%7D%20%5BCOOH%5D%5C%5C)
![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}}](https://tex.z-dn.net/?f=Rate%20%3D%20k_%7B4%7D%20%20%5Cfrac%7Bk_%7B2%7D%20%5BCO_%7B2%7D%20%5D%5BH%5D%7D%7Bk_%7B3%7Dk_%7B4%7D%20%20%7D%5C%5CRate%20%3D%20k_%7B4%7D%20%20%5Cfrac%7Bk_%7B2%7D%5BCO_%7B2%7D%5D%5Cfrac%7Bk_%7B1%7D%5BH_%7B2%7D%5D%20%20%7D%7Bk_%7B5%7D%5BOH%5D%20%2Bk_%7B2%7D%5BCO_%7B2%7D%5D%7D%20%20%7D%7Bk_%7B3%7Dk_%7B4%7D%7D%5C%5CRate%20%3D%20k_%7B4%7D%20%20%5Cfrac%7Bk_%7B2%7D%5BCO_%7B2%7D%5D%5Cfrac%7Bk_%7B1%7D%5BH_%7B2%7D%5D%20%20%7D%7Bk_%7B5%7D%5Cfrac%7Bk_%7B4%7DCOOH%20%7D%7Bk_%7B5%7DH%20%7D%20%20%2Bk_%7B2%7D%5BCO_%7B2%7D%5D%7D%20%20%7D%7Bk_%7B3%7Dk_%7B4%7D%7D)
Simplifying the equation above, the rate law becomes
![Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]](https://tex.z-dn.net/?f=Rate%20%3D%20%5Cfrac%7Bk_%7B1%7Dk_%7B4%7D%20%20%7D%7Bk_%7B3%7D%2B%202k_%7B4%7D%20%20%7D%20%5BH_%7B2%7D%20%5D)
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
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I hope this helps! Good luck!
Absolute positions — latitudes and longitudes
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False! Just saying. You could be under the influence, or just have no clue as to what you're doing.