I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti
1 answer:
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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
