Question

So, my chemistry homework is asking me to determine whether a reaction is spontaneous because of the increase of entropy in the system or in the surroundings. I have no idea what this means, can someone explain it please?

Answer 1

imagine putting some gas molecules in a corner of a closed room....what will the molecules do...spread throughout the room ,with time...did we do anything to cause it...no...so it was a spontaneous process...now entropy can be thought of as a measure of disorder or randomness in any system...first you had a fine room with gas molecules at some end...when molecules spread they caused disorder or randomness...hence if entropy of system is increasing,the process is spontaneous ,in general (thare is sometimes another factor named enthalpy-change involved)...

Answer 2

That at which your instructor is driving is the fact that all spontaneous processes increase the entropy of the universe. For example, in quarkine's example, the molecules of perfume in a newly opened bottle will spontaneously escape the bottle and drift around the room. This process, in which molecules initially confined to a small volume, escape into a larger volume, increases the entropy of the system. The reverse process, in which the gas molecules return to the bottle, would increase the entropy of the system.

However, such processes are not forbidden. A process that reduces the entropy of a system can still be spontaneous if at the same time the entropy of the surroundings (the rest of the universe) increases enough to compensate for the decrease in entropy of the system.

But how would that occur? With the transfer of heat. If heat is transferred from the system to the surroundings, then the entropy of the surroundings will increase. (You can see this through the Clausius statement of entropy, dS = dQ/T, where dS = change in entropy, dQ = heat added, T = absolute temperature) or by realizing that when you heat the surroundings the molecules in it speed up, and will generally have more freedom to move around.

And, indeed, if you were for example to cool the room in which the perfume molecules had escaped, transferring heat to the surroundings, then the perfume would eventually condense to a small pool of liquid. You would have decreased its entropy. But only by increasing the entropy of the surroundings by transferring heat out of the system.

In short, dS = dS_sys + dS_surr, where dS = change in entropy, sys refers to teh system and surr refers to the surroundings. For any spontaneous process, dS > 0. However, that only means the sum of dS_sys and dS_surr must be positive. It's possible for dS_sys to be negative (the entropy of the system decreases) so long as dS_surr is large and positive enough to compensate. When this happens, heat is always being transferred from the system to the surroundings.

To determine in each case whether a reaction is spontaneous because of dS_sys or dS_surr, you must look at two things: (1) whether it is endo- or exothermic. If it's endothermic, heat is being transferred INTO the system, which means dS_surr is negative, so this is clearly a case where dS_sys must be positive. (2) any factors about the reaction that tell you whether the reaction increases or decreases entropy. For example, if the reaction generates a gas from solid and liquid reactants, you know it increases entropy, because the entropy of a gas is always higher than that of a solid or liquid. If the reactants and products are all gases, and the moles of products are greater than the moles of reactants, this, too, increases entropy. If the products are in solution while the reactants are pure solids and liquids, this increases entropy (because of mixing). There are more cases, too.

Answer 3

Oops, sorry, the last sentence in my first para should read "The reverse process, in which the gas molecules return to the bottle, would decrease the entropy of the system."

Answer 4

Well, we have a list of reactions like
CO2(g) → CO2(s) @ -100°C
2I(g) → I2(g) @ 1000°C
MgCl2(s) → Mg2+(aq) + 2Cl-(aq) @ 25°C
KBr(s) → K+(aq) + Br-(aq) @ 25°C
CO2(s) → CO2(g) @ -50°C

and we have to select from a list either Enthalpy controlled, Entropy controlled, or Both... And no one I'm working with can figure it out. :-/

Answer 5

oh i forgot..i kinda gave the physics explanation..

Answer 6

All right, let's think them through, one by one:

(1) CO2(g) -> CO2(s) at -100oC. The sublimation point of CO2 is -78oC, so you know this process (condensation of the vapor to the solid) is spontaneous. But why? You also know that CO2 has modest intermolecular forces, because it has no H-bonding, no dipole-dipole forces, and only London forces. You also know the CO2 molecules are very far apart in the gas phase and very close together, and organized, in the solid phase. So going from the gas to the solid phase, two things will happen: first, there will be much more attractive (London force) intermolecular interaction between the CO2 molecules, so dH < 0, that is, this is an exothermic reaction -- heat will be released. Secondly, the entropy will drop considerably, because the entropy of the ordered solid is much less than the entropy of the gas. So dS < 0. Now, as we said earlier, dS_tot = dS + dS_surr MUST be > 0. Clearly since dS < 0 it must be dS_surr > 0. And that is possible, because indeed heat is released by the reaction (dH < 0), and this heat escapes into the surroundings and raises their entropy. In other words, this reaction is spontaneous only because the enthalpy of the CO2 falls so much the heat released increases the entropy of the surroundings enough to compensate for the reduction of entropy of the CO2. You can call this one enthalpy controlled.

(2) 2I(g) -> I2(g) at 1000oC. This is at a very high temperature (iodine boils at 184oC). Clearly when we combine the iodine atoms into the I2 molecule, and form the I-I bond, heat will be released (the I-I bond energy is 148 kJ/mol), so dH < 0 for this one. On the other hand, we are reducing the moles of gas from 2 to 1, so that will significantly decrease the entropy, dS < 0 also. From just the point of view of the signs of dH and dS, this is just like the CO2 case, and we have a bit of a competition between dH < 0 (which will raise the entropy of the surroundings, and tend to make the reaction spontaneous as written) and dS < 0, which will tend to make the reaction NOT spontaneous. Which will win? That's a little tricky. Keep in mind the Clausius formulation that tells you how much the entropy of the surroundings goes up for a given transfer of heat looks like this: dS_surr = Q/T, where Q is the heat transferred (in this case just -dH of the system, assuming everything happens at constant pressure). So that tells you the amount of entropy increase you get in the surroundings for a given dH < 0 of the system goes down as the temperature goes up. What that tells you is that at low enough temperatures, the dH < 0 fact of this reaction will dominate, and the reaction will be spontaneous. And, indeed, at low temperatures we certain expect to see I atoms spontaneously form I2 molecules. But at very high temperatures -- and perhaps 1000oC is high enough -- the dS_surr won't be large enough to compensate for the dS < 0 of the iodine, and the dissociation reaction will be spontaneous. You're probably best of answering"both" for this one. From a more advanced perspective, a quick way to estimate when this happens is to compare dH to RT (R being the gas constant), since RT is about the size of the molar entropy of a gas. If dH is bigger than RT, then the dH part dominates, and if RT is bigger, then the dS dominates. Here dH = -148 kJ/mol and RT = 10.6 kJ/mol, so even though the temperature is pretty high, dH will dominate and this reaction will be spontaneous as written.

(3) MgCl2(s) → Mg2+(aq) + 2Cl-(aq) @ 25°C Magnesium chloride is quite soluble in water, so unless you have a fully saturated solution already, this process is spontaneous. You will be tearing apart the MgCl2 crystal, meaning pulling Mg++ ions away from Cl- anions, and you know ion-ion forces are very strong, so this absorb a lot of heat. On the other hand, you are allowing Mg++ and Cl- ions to form strong ion-dipole intermolecular interactions with water moles, which will release the so-called heat of hydration of Mg++ and Cl-. That's pretty high, too. However, it's a good bet that the heat needed to pull apart the ionic crystal is much higher than the heat released by forming hydrated ions, so dH > 0 is a good assumption here. The entropy increases quite a bit, since you are going from an ordered crystalline solid to a disordered solution of ions. (The ions essentially form a gas within the water.) So dS > 0. In this case, then, the entropy of the system is decreasing but the entropy of the surroundings is going down -- because heat is being pulled out of them, to break up the crystal. So this is the opposite of the first two cases. Since the reaction is spontaneous, it must be that the entropy increase of the mag chloride is more than enough to compensate for the entropy decrease in the surroundings that comes from supplying heat. So this reaction is entropy driven.

(4) KBr(s) → K+(aq) + Br-(aq) @ 25°C Exactly the same situation. Not sure why this is also on here.

(5) CO2(s) → CO2(g) @ -50°C This is above the sublimation temperature, so you know this reaction is spontaneous as written. The analysis is the reverse of the first, meaning dH > 0 (the reaction absorbs heat, reducing the entropy of the surroundings) and dS > 0 (because you generate a gas from an ordered solid). Since the reaction is spontaneous -50, it must be the entropy increase of the CO2 is more than enough to compensate for the entroyp decrease of the surroundings. This is entropically driven, then.

In general, you'll notice to solve these problems you need to bring together a combination of theoretical concepts -- for example, the role of heat transfer in entropy, and how entropy depends on moles of gas, and what intermolecular forces exist between atoms and molecules -- and a microscopic (atom-level) understanding of what's going on, and empirical fact, for example what the melting point of a substance is, and knowing that above the mp melting is spontaneous, but below it freezing is. These are tough problems!