T-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be

Complete Question:

t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be

I + S ⇄ I*S

W + S ⇄ W*S

W*S + I*S ⇄ TBA * S * S

TBA * S ⇄ TBA + S

Derive a rate law assuming:

(a) The surface reaction is rate limiting

(b) The adsorption of isobutene is limiting

(c) The reaction follows Eley-rideal Kinetics

I*S+W ⇄ TBA * S

and surface reaction is limiting

(d) Isobutene (I) and water (W) are adsorbed on different sites

I + S₁ ⇄ I*S₁

W + S₂ ⇄ W*S₂

TBA is not on the surface, and the surface reaction is rate-limiting

(e) What generalizations can you make by comparing rate laws derived from part (a) through (d)?

Answer and explanation:

The mechanism for the production of t-butyl alcohol is as follows:

the reaction and rate law for the adsorption of isobutene over the amberlyst-15 is as follows:

I + S ⇄ I * S                                        

where is the concentration of vacant site

is the equilibrium constant of the adsorption

is the rate constant for forward

are concentration of isobutene and site filled with isobutene

The reaction and rate law for the adsorption of water (W) over the amberlyst-15 catalyst catalyst is as follows

W + S ⇄ W.S                                      

The reaction and rate law for the surface reaction on the catalyst is as follows

W.S + I.S ⇄ TBA . S + Sn                        

The reaction and rate law for the desorption of TBA from catalyst is as follows

TBA . S ⇄ TBA + S                            

the attached image below gives the remaining steps    

14,16 18 reacts with group

am i too late?

explanation: