Question

`(1)` T/F Do all lipids have a hydrophilic head and hydrophobic tail? `(2)` How is a rate constant related to the Gibbs' free energy of activation, `DeltaG^‡`?

Answer 1

1) Lipids may mean fats, waxes, sterols, fat-soluble vitamins, certain glycerides, phospholipids, and more.

As such, only some lipids have a hydrophilic head and hydrophobic tail (a phospholipid, for example). All literally means all, without exception.

A phospholipid looks like this:

https://upload.wikimedia.org/

The head group is made up of a phosphate and some other component, while the tail is basically a derivative of a two-tailed fatty acid (a carboxylic acid with a long alkyl tail).

http://usercontent1.hubimg.com/

But again, not all lipids have this characteristic. Some are simply hydrophobic but not hydrophilic.

2) The rate constant `k` is actually proportional to the Gibbs' free energy of the transition state for a reaction:

`\mathbf(k prop e^(-color(green)(DeltaG^‡)"/"RT))`

http://chemwiki.ucdavis.edu/

...kind of like how it is proportional to the activation energy of the reaction:

`\mathbf(k = Ae^(-color(green)(E_a)"/"RT))`

https://upload.wikimedia.org/

Catalysts actually tend to somehow alter the mechanism of the reaction to lower the activation energy, making a reaction kinetically favorable.

A common way to do this is to stabilize the transition state. Enzymes are frequently said to alter `DeltaG^‡`, thus altering `k` as well.

If you increase `["Catalyst"]` and the catalyst is supposed to target and stabilize the transition state, then `DeltaG^‡darr`, hence `k uarr`. That correlates to the idea that a more stable transition state has a lower Gibbs' free energy of the transition state and a lower activation energy.

A higher rate constant thus results, which makes sense because a catalyst is supposed to speed up a reaction. A higher `k` corresponds to less/faster time, since rates are `"something/s"`.