Question

How can a thermodynamically unfavorable reaction be made thermodynamically favorable?

Answer 1

A. By coupling it with a series of favorable reactions such that the reactions add up to produce an overall reaction with G° = 0.
B. By coupling it with a series of unfavorable reactions such that the reactions add up to produce an overall reaction with a negative G°.
C. By coupling it with a series of favorable reactions such that the reactions add up to produce an overall reaction with a positive G°.
D. By coupling it with a series of favorable reactions such that the reactions add up to produce an overall reaction with a negative G°.

Answer 2

@Jesstho.-.

Answer 3

Use this formula

G = H + change in temperature*entropy

Answer 4

What do you think? @vera_ewing

Answer 5

D right?

Answer 6

Hm, hang on.

Answer 7

ok...

Answer 8

Basically, it can take a lot of activation energy to make a thermodynamically favorable reaction happen. That's the simple version, but I'll go into more detail to show what underlies this "simple" statement:

It's thermodynamically favorable for you to burst into flames right this second and become a pile of ash and hydrogen/oxygen gas. But this doesn't happen. Why not?

When we talk about thermodynamics, what we're really talking about is statistics. Statisically, it's extremely unlikely for millions of atoms to come together all in one place and stay still. It's very likely for them to be scattered around randomly. When we say "thermodynamically favorable" we just mean "more likely". If you put a drop of ink in a glass of water, it spreads around the glass evenly. There's nothing in the laws of physics that says it won't eventually, randomly go back to being just a drop. It's just extremely unlikely. So the state of being 'spread around' is favorable (more likely)

So, to take your paper example: It's very unlikely for a bunch of carbon and hydrogen atoms to be sitting around in one small area like that. It's very likely for them to be scattered in the wind, as would happen if they burst into flames. So thermodynamically speaking, the reaction of paper bursting into flames is favorable.

However, even a favorable reaction has to *actually happen*. Turns out this is often the hard part. While the carbon and hydrogen is in a thermodynamically unfavorable state, it's also bound there by some pretty strong covalent bonds. To break them, the oxygen must hit really, really hard. And, at room temperature, oxygen just doesn't have enough kinetic energy to break these bonds. (Remeber, "temperature" is just another way of saying "average velocity (energy) of the molecules" - a high temperature means more molecules going faster).

Now, at higher temperatures, two things happen.

First, the carbons and hydrogens in the paper will be jiggling back and forth more, straining their bonds. Second, the oxygen molecules will be flying around faster. So, now, maybe one or two oxygens DO have enough speed to hit one of those bonds and break it. And when that happens, the molecule will start to break down into its thermodynamically favorable state.

So this is why some thermodynamically favorable reactions do not happen at room temperature, but do if you increase the heat. The reaction will be able to overcome that high activation energy.

We have an equation for this, and you may have seen it:

G = H - T*S

T is temperature.

S is how "thermodynamically" favorable the reaction is.


H how much energy is stored in our bonds we have to break.

When G is less than 0, the reaction occurs spontaneously. The larger G is, the faster it will happen.

You can see that as temperature goes up, we are more able to break "unfavorable" bonds.
Via Yahoo :P

Answer 9

:3

Answer 10

So what is the answer? Is it C?

Answer 11

I'd say.

Answer 12

thanks :)

Answer 13

Read what @Jesstho.-. said. Try to find the final answer on your own :)