rate constant

[bluequestions]

I understand that according to the Arrhenius equation, k = A (e^-x/T), where x = Ea/R, when you increase T, the exponential term just becomes larger, so the rate constant increases, and this is true for both forward and reverse reactions. However, I was also wondering if we designate the rate constant of forward reaction as k, we often designate the reverse rate constant as k^-1 (or 1/k), wouldn't increasing the forward rate constant k lead to a smaller reverse rate constant though? Doesn't this contradict what happens when we estimate change in rate constants using Arrhenius equation?

Thank you!

---

Members don't see this ad.

 

[sazerac]

If I get what you are asking, the reverse reaction rate for the first reaction is often called k with a SUBSCRIPT of -1, not a superscript. The -1 is meant to be just a label indicating backwards 1, and is not meant to imply subtracting 1 from k or raising k to a power of -1.

Many examples use kforward and kreverse instead of 1 and -1, to avoid any confusion.

 

[aldol16]

I understand that according to the Arrhenius equation, k = A (e^-x/T), where x = Ea/R, when you increase T, the exponential term just becomes larger, so the rate constant increases, and this is true for both forward and reverse reactions. However, I was also wondering if we designate the rate constant of forward reaction as k, we often designate the reverse rate constant as k^-1 (or 1/k), wouldn't increasing the forward rate constant k lead to a smaller reverse rate constant though? Doesn't this contradict what happens when we estimate change in rate constants using Arrhenius equation?

I think you're confusing rate constants with equilibrium constants here. Equilibrium constants have the K(forward) = 1/K(reverse) relationship. Rate constants don't necessarily have that. In fact, for a system at equilibrium, K(eq) = k(forward)/k(reverse). So if forward rate constant increases, so does reverse rate constant as long as that system is at equilibrium.

Now, keep in mind also that the activation energy is not the same for forward and reverse reactions as long as the reaction is either endothermic or exothermic.

 

[bluequestions]

Got it! Thank you to both of you!