Imagine two competing first-order reactions with rate constants k1 > k2. In othe

Question

Imagine two competing first-order reactions with rate constants k1 > k2. In other words, the first reaction is faster. Show mathematically that, as you lower the temperature, the selectivity will normally increase. Hint: Selectivity can be measured as the ratio between two rate constants. A larger ratio of rate constants almost always gives a more selective reaction. You can use the relationship between k and G ‡ to make this proof. Which equation defines the relationship between k and G ‡ ? (15 points) Bonus: Why is this only an approximation? Under what conditions can the selectivity remain temperature-independen

Explanation / Answer

The rate of reaction is inversrly protortional to the temperature of the reaction.

This is according to the arrhenius equation,

lnk = -Ea/RT + lnA

So as the temperature is dereased, the rate of reaction and hence the selectvity would increase over time.

The relationship between k and dG is,

dG = -RTlnk

where, dG is gibbs free energy

R = gas constant

T = temperature

At very high k value the selectivity remains independent of temperature.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.