rate constant versus rate of reaction...im a bit confused

[goldngrl1611]

Hey quick question for you guys-
I understand there is a difference between the rate constant, k, and the rate of reaction. A catalyst, for example, only affects the rate of reaction, but not the rate constant, which is only affected by temperature. But, what do these tell us about different rates? In other words....

A high rate constant means what? A quick rate at which you reach equilibrium, or a fast rate at equilibrium at which your reactants and products are forming and reforming? Then the same thing for a rate of reaction: is this the rate your reaction proceeds to equilibrium, or the rate your products and reactants are forming/reforming once it is reached?

If someone can differentiate the two for me it would be super helpful. Thanks!

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[SoftwareKevin]

Hey quick question for you guys-
I understand there is a difference between the rate constant, k, and the rate of reaction. A catalyst, for example, only affects the rate of reaction, but not the rate constant, which is only affected by temperature. But, what do these tell us about different rates? In other words....

A high rate constant means what? A quick rate at which you reach equilibrium, or a fast rate at equilibrium at which your reactants and products are forming and reforming? Then the same thing for a rate of reaction: is this the rate your reaction proceeds to equilibrium, or the rate your products and reactants are forming/reforming once it is reached?

If someone can differentiate the two for me it would be super helpful. Thanks!

I think a catalyst *does* change the rate constant. If I remember right, a catalyst will lower the activation energy and the Arrenhius equation is:
k = A*e^(-Ea/RT)

Based on the above formula, lowering the activation energy should give us a larger rate constant.

The rate of reaction will change as the reaction proceeds towards equilibrium because rate of reaction is dependent on concentrations. As we move towards equilibrium the forward rate will slow down and the reverse rate will speed up. At equilibrium the forward rate is equal to the reverse rate of reaction.

Rate of reaction is directly related to the rate constant. So the larger the rate constant, the faster the reaction (both before you're at equilibrium and when you're at equilibrium)

 

[Phantastic]

The above poster is 100% correct. The thing you are trying to distinguish, or may have been told to distinguish in the past is:

Catalysts act kinetically, not thermodynamically.

Thermodynamics = gibbs, enthalpies, K values, pK values...essentially anything that denotes the concentrations of products and reactants at equilibrium.

Kinetics = rates, temperature, activation energies, etc...essentially anything that affects how quickly we reach equilibrium.

e.g. a diamond is always turning into graphite, but very very slow due to high activation energy. The process of turning to graphite is said then to be favorable (thermodynamics), and the process to get there is slow as all hell (kinetics).

Easy to confuse the word "favorable" with how quick it reaches equilibrium, so AAMC just loves to eat up your points that way!