Need help intuiting a few chemistry concepts (rate equation and equilibrium)

Factual Questions
1

Unfortunately I’m only learning from Crash Course Chemistry on Youtube where every lesson is 10 minutes long but even after searching for quite a while on Google, I’m having trouble getting an intuitive explanation for the questions below. I have a feeling I’m going to be iminently saying “duh.” Here goes:

  1. Why does the stoichiometric coefficient become an exponent in the rate equation (for an elementary reaction)? I would think if you double the amount of a reactant it would double the reaction rate since there are twice as many molecules to collide with. Why does it become an exponent?

  2. Why does increasing the pressure of a reversible reaction favor the the side with fewer moles? I understand that at a lower volume, more collisions will occur, but isn’t it just as likely those increased collisions will break apart an equal amount of the lower mole products? I get that if you increase the pressure and keep volume constant then temperature must increase so I’d think that would favor the endothermic direction. The only explanations I get is that the system “attempts to undo what you just did” which drives me up a fucking wall.

Thanks in advance and don’t forget I’m looking to understand this stuff intuitively.

2

I’ll address Question #1 first.

Caveat: all of the following only applies to elementary (single step) reactions.

For reaction A + B ----> products, the forward reaction rate is given by:
rate = k [A] **

So if you double the concentration of either A or B, you double the reaction rate, because there are now twice as many collisions between reactant molecules. If you double the concentration of both, you quadruple the reaction rate, because there are now four times as many collisions between reactant molecules.

Continuing on, for reaction A + B + C ----> products, the forward reaction rate is given by:
rate = k [A] ** [C]

Finally, for reaction A + 2B ----> products, the forward reaction rate is given by:
rate = k [A] ** ** = k [A] **[sup]2[/sup]

So as you can see, that’s why the stoichiometric coefficient becomes an exponent in the rate equation.

3

The second one seems obviously intuitive to me. Consider a simple case with two moles on the left and one mole on the right,
A + B <> C
The forward reaction requires a collision between A and B. The reverse reaction, the decomposition of C, requires no collision at all. Since only the rate of the forward reaction is affected by changing the collision rate, increasing pressure pushes equilibrium right. Conversely, imagine an extreme case where the pressure is so low that molecules never collide. Decomposition of C would still occur, but the forward reaction would cease - equilibrium would be fully left, no C at all.

4

Question #2: First off, note that your example here only applies to gaseous equations.

Consider reversible reaction A (g) + 2B (g) <----> 2C (g)

There are three moles on the left side of the equation, and two moles on the right side of the equation. As in all reversible reactions at equlibrium, the forward reaction rate at equilibrium is exactly equal to the reverse reaction rate. If you increase the pressure, you increase the forward reaction rate, because there are more collisions between reactant molecules. The reverse reaction rate increases as well but not as much as the forward reaction rate. That’s because the forward reaction rate is given by:

rate[sub]forward[/sub] = k P[sub]A[/sub] P[sub]B[/sub][sup]2[/sup]

While the reverse reaction rate is given by:

rate[sub]reverse[/sub] = k P[sub]C[/sub][sup]2[/sup]

(Note that for a gaseous reaction, we use partial pressures instead of concentrations for the rate equations.)

So if you double the pressure, the forward reaction rate increases by eight times (2 x 2[sup]2[/sup] = 8), while the reverse reaction rate only increases by four times (2[sup]2[/sup] = 4).

Because the forward reaction rate is now greater than the reverse reaction rate, this has the effect of increasing the amount of products, which therefore increases the reverse reaction rate until it again matches the forward reaction rate. You end up at a new equilibrium with the reaction rates being equal again, but with a higher concentration of C than before (and a lower concentration of A and B).

P.S. I wouldn’t bring temperature into it. That has an effect as well (i.e. favoring the direction that is endothermic in the case of increasing temperature), but that’s just confusing the issue.

5

P.P.S. For reactions in solution, such as aqueous reactions, you use concentrations for rate equations. The symbol used is: [A] = concentration of A (mol/L)

For gaseous reactions, you use partial pressures. The symbol used is: P[sub]A[/sub] = partial pressure of A (measured in atm, torr, kPa, etc.)

Other than this difference in terminology, they both work the same in rate equations.

With that said, I initially wrote my last post using concentrations, then edited it to change them to partial pressures (to be more correct for a gaseous reaction). Upon re-reading my post, i looks like I missed a few, though. Also, I should have used k[sub]P[/sub] for the rate constant.

Regardless, not too bad to do all of this from memory, if I don’t say so myself. I haven’t looked at this stuff in years! However, like I’ve told others, if you want to learn something better than you thought possible, become a teacher. In my case, I taught this material to 3-4 classes a year for five years. That was 20 years ago, though…

6

I was working under the assumption that the decomposition of C would require the energy from a collision for its activation energy. That either an A, B, or C would have to collide with C to knock C apart. If that’s not the case then it is intuitive. But what provides the energy for C to decompose? Thanks.

7

I’m with you right up until A+2B uses 2 as the exponent for B. Like you said, if you double the reactants, you double the reaction rate, but using the exponent makes the reaction rate proportional to the square.

8

Temperature is a measure of average energy. At any given temperature, the individual molecules have a distribution of energy. So some have more and some have less.

9

I’m rapidly going to get out of my depth here, but I think correct basic intuition here is that the significance of the kinetic energy in a collision is that it overcomes electrostatic repulsion to bring things closer together.

10

Depending on the type, Molecules have three degrees of freedom : Translational (like balls bouncing around), rotational (when two or more atoms join , they can spin around their center) and Vibrational (if two balls are joined together by a spring, they can vibrate ).

When you increase the temperature of a gas, the energy can go into feeding all these degrees of freedom. So as you can see, since C is at least diatomic - it will vibrate more or spin faster and hence can break the bonds holding it together.

11

But I think introducing temperature variation is a separate matter. Assume constant temperature, and vary the frequency of collisions by increasing pressure. That seems to be the situation that OP is seeking intuition for under collision theory.

12

Go back to reaction A + B + C ----> products.

The forward reaction rate is given by: rate = k [A] ** [C]

In this reaction, if you double the concentration of either A, B, or C, you double the reaction rate. If you double the concentration of two of the three reactants, you quadruple the reaction rate, and if you double the concentration of all three reactants, you increase the reaction rate by eight times (2 x 2 x 2 = 8).

Moving on, Reaction A + 2B ----> products is the same as A + B + B ----> products. In either case, the forward reaction rate is given by:

rate = k [A] ** ** = k [A] **[sup]2[/sup]

In this case, if you double the concentration of A, you double the reaction rate. But if you double the concentration of B, you increase the reaction rate by four times. This is because B is involved twice in the reaction. Increasing its concentration therefore has a double effect, which is reflected by the exponent in the rate equation.

13

Oh but I thought the OP started talking about the activation energy, which is temperature dependent.

So for the reaction, A + B —> C, Activation energy (which depends on temperature) is not changed by the pressure. You may write the equation as:

Rate of disappearance of C = A * exp(-E/RT) * [C]

Where [C] is the concentration or partial pressure. As pressure goes up, the concentration or partial pressure of [C] goes up and the reaction rate increases.

Rate of formation of C also goes up in the same manner. But at the higher pressure, the equilibrium constant (not the reaction rate) favors C

I think the OP is confusing reaction rates to equilibrium constant.

14

So you’re saying the moles/liter value used for B in A+2B -> C is only for 1B (and therefore must be included twice)?

15

Well if you increase the pressure and the volume is kept constant, then the temperature goes up, no?

16

Well that makes alot of sense if it’s applicable to pressure as well, or I’m right about the interpretation of gas law in my previous post. Which I’m probably not.

17

We are talking about fixed number of moles of a gas.

So, you can keep volume constant, increase the temperature and the pressure will go up. Like a gas in a sealed can.

18

KidCharlemagne - I think you will benefit from distinguishing between reaction kinetics, reaction equilibrium and the kinetic theory of gases. Let’s just take a look at reaction kinetics and reaction equilibrium in this example:

Say Red molecules react with Blue molecules to form Green molecules.
Red + Blue <–> Green

The forward rate of reaction, i.e. the rate of formation of Green molecules (which is also the rate of disappearance of R and B) is given by r_forward = kfPP_RedPP_Blue. Now kf (k forward reaction) is a function of temperature and temperature only. PP implies partial pressure.
The reverse rate of reaction, i.e. the rate of decomposition of Green molecules (which is also the rate of reappearance of R and B) is given by r_reverse = kr*PP_Green. Again Now kr (k reverse reaction) is a function of temperature and temperature only. PP implies partial pressure.

So let’s take a vessel maintained at a constant temperature (so kf and kr don’t change) and put in 30 of Red, 30 of Blue and 40 of Green. Lets say kf = 0.01 molecules/min per (psi^-2) and kr = 0.1 molecules/min per (psi^-1). Lets say that the pressure at the start is 100 psi and therefore the partial pressure of Red is 30 psi (PP_Red =30 psi ), the partial pressure of Blue is 30 psi (PP_Blue = 30 psi) and PP_Green = 40 psi.

Although, this is a differential equation, lets look at minute by minute.

**Start : 30 Red 30 Blue 40 Green and total pressure of 100 psi **
In the first min, r_forward = 0.01 molecules/min per (psi^-2) * 30psi * 30psi = 9 molecules of Green are formed
Also, in the first min, r_reverse = 0.1 molecules/min per (psi^-1) * 40psi = 4 molecules of Green are broken up.
End of 1 min, we will have Red (30 -9 +4 ) Blue (30-9+4) and Green (40+9-4)

** Start of 2 min: 25 Red 25 Blue 45 Green and total pressure of 95 psi **
In the second min, r_forward = 0.01 * 25psi * 25psi ~ 6 molecules of Green are formed
Also, in the second min, r_reverse = 0.1 * 45 ~ 5 molecules of Green are broken up,
End of 2 min, we will have Red (25 -6 +5), Blue (25 -6 +5) and Green (45+6-5)

** Start of 3 min: 24 Red 24 Blue 46 Green and total pressure of 94 psi **
In the third min, r_forward = 0.01 * 24psi * 24psi ~ 6 molecules of Green are formed
Also, in the third min, r_reverse = 0.1 * 46 ~ 5 molecules of Green are broken up,
End of 3 min, we will have Red (24 -6 +5), Blue (24 -6 +5) and Green (46+6-5)

** Start of 4 min: 23 Red 23 Blue 47 Green and total pressure of 93 psi **
In the third min, r_forward = 0.01 * 23psi * 23psi ~ 5 molecules of Green are formed
Also, in the third min, r_reverse = 0.1 * 47 ~ 5 molecules of Green are broken up,
End of 4 min, we will have Red (23 -5 +5), Blue (23 -5 +5) and Green (47+5-5)

** At the end of 4 mins, equilibrium is reached. There are constant 23 Red 23 Blue 47 Green molecules and total pressure of 93 psi. Every minute, 5 Green are newly formed and 5 Green disintegrate back **

Now suppose, the pressure is increased from 93 psi to 130.2 psi (40% increase) keeping temperature the same. The partial pressure of Red is 32.2psi, Blue is 32.2psi and Green is 65.8psi.
**Start : 23 Red 23 Blue 47 Green and total pressure of 130.2 psi **
In the first min, r_forward = 0.01 molecules/min per (psi^-2) * 32.2psi * 32.2psi ~ 10 molecules of Green are formed
Also, in the first min, r_reverse = 0.1 molecules/min per (psi^-1) * 65.8 psi ~ 7 molecules of Green are broken up.
End of 1 min, we will have Red (23 -10 + 7 ) Blue (23 -10 + 7) and Green (47+10 -7)

** Start of 2 min: 20 Red 20 Blue 50 Green and total pressure of 126 psi **
In the second min, r_forward = 0.01 * 28psi * 28psi ~ 8 molecules of Green are formed
Also, in the second min, r_reverse = 0.1 * 70 psi ~ 7 molecules of Green are broken up,
End of 2 min, we will have Red (20-8+7), Blue (20-8+7) and Green (50+8-7)

** Start of 3 min: 19 Red 19 Blue 51 Blue and total pressure of 124.6 psi **
In the second min, r_forward = 0.01 * 26.6psi * 26.6 psi ~ 7 molecules of Green are formed
Also, in the second min, r_reverse = 0.1 * 71.4 psi ~ 7 molecules of Green are broken up,
End of 3 min, we will have Red (19-7+7), Blue (19-7+7) and Green (51+7-7)

** At the end of 3 mins, equilibrium is reached faster than at lower pressure. There are constant 19 Red 19 Blue 51 Green molecules and total pressure of 124.6 psi. Every minute, 7 Green are newly formed and 7 Green disintegrate back **

So, I hope you see that just the rate equations make the reaction equilibrium shift to the right (more Green) at higher pressure (because the partial pressure term is squared in the forward reacton). Also note that the rate of reaction is higher at the higher pressure both for the forward and the reverse reaction.

19

Thanks for that detailed response. I certainly see how it works out mathematically, but its when I try to think of it in terms of how the increasing number of collisions favors the higher mole reactants that my intuition breaks down. If you assume (and this may be a bad assumption) that anything colliding with the diatomic lower mole product favors decomposition, while 2 higher mole reactants colliding favors creation of product, when I add up the potential permutations/combinations of collisions, it seems to favor both equally. Not sure if my math or assumptions about how collisions affect reaction rates is the culprit in my misunderstanding.

20

I think this is where we need to examine your intuition. The increased pressure effects all reactions in the same way I.e. it makes the reactions faster. If you see above in the example red + blue is reacting faster to make green. And green is breaking up faster to make red + blue.

The question you need to ask your intuition is what will happen at higher pressures wrt equilibrium? If the forward and the backward reaction both go up in speed, with higher pressure - where will it balance out ?