Equilibrium and Entropy

Hey Alex, 

I've been having trouble with some general chemistry material and I was using the Q-bank quizzes when I came upon the following question. 


14. Reaction 1 proceeds in a closed flask and reaches equilibrium. After a while, the flask beings to leak NO (g). Once the leak is fixed and the reaction has once again reached equilibrium, the entropy of the system has:

a. increased
b. decreased
c. remained constant
d. decreased by the same amount as delta g

reaction 1: 3 NO(g) --> N2O(g) + NO2(g)

The answer is a. I don't understand why. If the reaction has once again reached equilibrium, then how has the entropy changed? Is it because there was a net increase in entropy to once again reach equilibrium? How should we go about doing this?
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Initially, you may want to attack this question using ΔG = ΔH-TΔS.  ΔG = 0 at equilibrium, by definition.  There is a problem with making an assumption that ΔS will have to stay the same, though -- T will likely not stay the same (if some of the NO is leaking, the pressure inside goes down and the gas will expand to fill the container; an expanding gas will lose temperature, according to the First Law of Thermodynamics (ΔU = Q - W)).


So what can we do?  Well, let's think about LeChatelier's Principle.  When the NO escapes, the system is no longer at equilibrium.  Once we fix the leak, let's think what will happen to the system.  With a lack of NO, the system will shift left.  Remember that entropy is increased in gases (over liquids and solids) and that more moles of gas corresponds to more entropy.  This left shift, therefore, is increasing entropy as the reaction reaches equilibrium.


I hope this helps!

Translational motion test

Hi I have questions on the physics Translation Motion Test: 

Please note that I answered some of these questions elsewhere in the blog, but I'll re-answer them here.

2. Terminal velocity? how is it we can assume A=0 and no net forces. ?? So every object has a terminal velocity where A=O = gf so it does not accelerate further???
Terminal velocity, by definition, is the point when the drag force is equal to the weight.  With two equal and opposite forces, that object will experience no acceleration, as perNewton's Second Law (F=ma).  Every object that has a drag force acting on it will eventually reach terminal velocity if it keeps falling.  Once it does, the acceleration is zero.  That doesn't mean it's not moving; it just means the velocity is constant.

4. q = (KLsquared) / (Kprime L cubed) = ((K) / K prime) (1/L)
UM don't get the creation of K and Kprime, are we making the values of the proportionality constant up? And why wouldn't they be the same
The invention of K as a proportionality constant is a mathematical convention, but you really don't need it here.  If q is surface area/volume, well, they tell us that surface area is proportional to L² and volume is proportional to L³.  So, q would be proportional to L²/L³, or 1/L.  Adding in a proportionality constant is just a convention.  From there, q-ant/q-human would be (1/L_ant)/(1/L_human).  Ugly fraction, so let’s invert and multiply:  q-ant/q-human = L_human/L_ant.  L_human it says is 1 m and L_ant is 0.5 cm, or 5x10^-3m.  L_human/L_ant is then 1 m/(5x10^-3 m) = 0.2x10³ = 200.  The ratio is therefore 200:1.

5. In Q5 we drop the assumed K we used in 4 - How/why/ When is it appropriate to create these assumptions? - then we assume the length of the daughter cell keeps the same L squared and L cubed relationships with out the proportionality constant....
Then they create the formula L squared = (2 raised the the 2/3) wavelength squared ( I know its not a wavelength but thats what the symbol looks like :) ) where the second portion of the equation = surface area of the original cell, since that is the case and its what we are starting with then HOW can you go forth and say 2wavelenght squared = surface area of daughter cells and NOT use what I thought it would be which is (2 raised to 2/3 ) times wavelength squared.   Then my inability to do math pops up (despite reviewing the math foundation)
2:2 raised 2/3 = (2) / (2 raised 2/3) = which apparently = (2)/(2 raised 2/3) times (2 raised to the negative 2/3) / (2 raised to the negative 2/3) = 2 raised to the 1/3 / 1 = which they then changed back into a ratio? = 2 raised 1/3 : 1
After this question I was ready to freak out. I just don't get it.

This is one of the most asked about questions in the whole Kaplan curriculum.  Here’s my explanation:

This question reads:  “Cells can only absorb nutrients through their surfaces.  In order for a cell to absorb nutrients most efficiently, it will grow to a maximum size and then divide.  If the total volume remains constant when a cell divides, then the ratio of the total surface area of the daughter cells to that of the original cell is:”

Let’s start with defining a few variables here.  According to our passage, “each animal has a characteristic length L…its area is proportional to L²…and its volume is proportional to L³.”

Let’s also define the subscript “1” to mean the cell before it’s divided, and “2” to represent each cell after it’s divided.

The total volume here is constant – the one cell before division has the same volume as adding together the two cells after division.  This gives us the equation that L₁³=2L₂³.  Let’s take the cube-root of both sides.  Well, the cube root is the same as taking each side to the 1/3 power.  So we would have:

(L₁³)^1/3=(2L₂³)^1/3.  When you have an exponent raised to an exponent, we multiply their values.  This means that this simplifies out to L₁=2^1/3L₂.  The 2^1/3 is because we have to distribute the exponent to both values in the parentheses.

From here, we can figure out the answer to our question.  It asks us to find the total surface area of the daughter cells, as compared to the area of the original cell, or

2L₂²:L₁².  Remember that there are TWO daughter cells, so that’s why we have the two here.  Well, we know how L₁ and L₂ are related, so we can say that

2L₂²:L₁² can be represented as 2L₂²:(2^1/3L₂)² – we just substituted in our value for L₁ from that volume calculation before.  Distributing our exponent on the right-hand side, we have 2L₂²:2^2/3L₂².  Let’s divide both sides by L₂².  That gives use 2:2^2/3.  But… that doesn’t quite look like any of the answers here.  Since most of the answers have 1 on the right-hand side, let’s divide both sides by 2^2/3.  That gives us:

[2/2^2/3]:[2^2/3/2^2/3].  When we divide two items that have the same root and different exponents, we simply subtract the exponents.  That means that we have [2^1-2/3]:1, or 2^1/3:1.  That corresponds to answer choice (C).  If you’re having difficulty with exponents, I encourage you to go on to Google and type in “Exponent Practice.”  You want to be able to work with exponents very easily on the MCAT.  The four big rules are:

(X^A)(X^B) = X^A+B

(X^A)/(X^B) = X^A-B

(X^A)^B = X^AB

nth root of X = X^1/n

6. How is it A=mg in the Y direction or is this only when calc  Force = (M) (-g) 

Acceleration will always be in the same direction as the net force on the object (remember F = ma).  So, the ONLY force on this object is gravity.  Since gravity points downwards, our acceleration is downwards.  Also, if that’s the only force, then F = mg = ma.  So a = g.  But we know that.  “g” is the acceleration due to gravity by definition.

14. I doubled the M in Ke = mv squared /2 but that was wrong since I did adjust for delta V with less mass BUt V = DELTA X / DELTA T
right? but when V initial = F delta T / M

This is what most people do at first with this question.  But you need to keep in mind when more than one variable may get affected.  If you have a heavier clown being pushed by the same force, he’ll move slower.  The method they give for answering this question – what you allude to at the end of your question here – I think is a bad way of approaching it. Where do they talk in the passage about the relationship between mass and velocity?  The graph.  Now, the y-axis on the graph is v_y and they’re asking in the question about v₀. But v_y and v₀ are proportional (it’s always the same angle that the clown gets fired at, so v_y/v₀ is a constant (it’s equal to sinθ here – opposite over hypotenuse).  So, let’s look at what happens when you double the mass.

Starting with the 40 kg line, it shows us that the initial velocity is 42 m/s.  If we double the mass to 80 kg, initial velocity is 21 m/s.  In other words, doubling the mass halves the velocity.  Since K = ½ mv², doubling the mass will cause m to double and v to get cut in half.  The net effect?  We get cut in half for our kinetic energy.

Re-Cap of Gen Chem III

Hey everyone!

Your Required Homework before Verbal Reasoning III is as follows:

• Chapter 9, question 2 in Writing Sample Review Notes
• Critical Reading Challenge Workshop & Quiz
• Argument Dissection Challenge Workshop & Quiz

For those of you who would like some additional Verbal Reasoning review before class, I recommend the following:

• Verbal Reasoning Foundation Review Unit 3

Moving on, let’s discuss the review assignments for Gen Chem III:

• Acids & Bases Test 1
• Electrochemistry Test 1
  • Both of these topical tests serve as a nice recap on the concepts we discussed in the lesson.  There are a few questions that get carried away with the math, but the exams get the job done: they serve as a nice test on all the general chemistry you need to know for the exam.

Content Note:  A real-world application of redox reactions is demonstrated in this “explosive” YouTube video.  A violent redox reaction occurs when you put an alkali metal and water together as electrons are transferred from the metal to water.  Alkali metals are potent reducing agents, and this comes at no surprise given the low ionization energy of metals.  The other reactant, water, is the oxidizing agent as the hydrogen in water (+1 oxidation state) gets reduced to hydrogen gas (0 oxidation state).  So why the explosion?  When H₂ and O₂ hook up we get a combustion reaction, and there’s plenty of O₂ in the atmosphere.  (Side note: the glass shatters because the product of the redox reaction, CsOH, is a strong enough base to etch glass).  So this one tiny 25-second video combines concepts we discussed in Gen Chem I (ionization energy), Gen Chem II (combustion reactions), and Gen Chem III (redox reactions)!      

Helpful Hints:

• Now that we are in the 3^rd Unit of the Kaplan MCAT course, our focus is on integration of all our methods and strategies in the context of timed sections.  From this point on, make sure that you complete all Section Tests and Full Length Practice Tests under timed conditions.  In addition to practicing your pacing, you must work on building stamina and focus in anticipation of Test Day.
• The Gen Chem III lesson applies many of the same concepts we discussed in Gen Chem II.  Remember that the dissociation constant of an acid or base (K_a and K_b, respectively) is a measure of the equilibrium position in the same way that K_eq and K_sp are.
• Concentration cells are like voltaic (galvanic) cells in that they house a spontaneous redox reaction which releases energy.  Unlike voltaic cells, however, concentration cells have a standard EMF of 0 V because the redox reaction is “motivated” not by a difference in reduction potentials between two chemically different electrodes but rather by a concentration gradient of an ionic species (e.g. the proton-motive force across the inner mitochondrial membrane – the mitochondrion IS a concentration cell/battery!)

Also, here’s an interesting link.

• The Big Picture: fascinating pictures from the world of microscopy.  Those of you who like picture books may enjoy this blog, because all they do is post big pictures with scant explanation.

Density of Gases

Hey Alex,

One of the questions asked: which of the following is true concerning a gas? The answer was Gas molecules generally move so fast and for such short distances that the change in their velocities due to gravity has a negligible effect on their behavior.

For this question, think about what gravity would do to each individual gas molecule -- it's trying to pull them down towards the ground.  How much of an effect will this have?  Well, the amount of time between collisions in the gas particles (and with walls of the container) is really really small (probably on the order of thousandths of seconds... at most).  Over than incredibly short time, how much change in velocity will these particles really experience?  Honestly, very little.  This should also just make intuitive sense to you -- air doesn't tend to "settle" in a room; granted, there is turbulence and all sorts of other factors, but the idea still remains.  Ultimately, the effect gravity has on these particles has basically as much impact as things like small eddys and currents that may occur from heat differences throughout the gas, etc.  Not much of an effect overall.

But then a question later asked: which of the following phases experiences the greatest change in density due to gravity: The answer was gas.  I'm not sure I understand this, I know gas is the most compressible, but isn't gravity supposed to not affect it that much? 

Understandably, this is somewhat hard to reconcile with the previous question.  But the question doesn't say which one has a significant change in density, just which has the most of the various phases.  Since gases are the most compressible, they'll have the largest change in density (but the absolute change here will be on the order of like 10^-20 -- I made up that number, but hopefully you get the idea).

Also..I think I confuse how to understand density because I know density of an object is an intrinsic property... then why is it (another answer to a question) that if you have two rooms connected by a closed door and you open the door - the mass and energy change but not density and pressure? I thought that you would have to use rho = mass/vol. And the answer says that density is an intrinsic property of gas... but that implies that density won't ever change won't it? I guess my question is.. when do we think of density as an intrinsic property and when can we think of it as a changing variable in a problem?

The definition of the word "intrinsic" in physics means something that does not depend on the amount of material you have.  In other words, 1 L of a gas with a density of 1 gram per cubic meter has the same density as 100 L of that same gas.  Other intrinsic properties include molarity, molality, specific heat, etc.  An extrinsic property is one that depends on amount, such as mass, volume, or heat capacity.  Thus, density is an intrinsic property but you can always change it.  Basically, "intrinsic" does not mean the same thing in physics as it does in common English usage.

More gen chem questions

1. (pg 57, Lesson Book): Would you post the definition for an informational passage (e.g. structure, content, etc.), the things to notice (either for our maps or in general), and the common types of MCAT questions which follow? Similarly, could you respond to the previous question for the other two passage types? [Although you stated them in class, I wanted to have a clear approach to each equation type]

Informational - presents info like a textbook; look for variable relationships, new terms, new definitions.  Questions tend to focus on variable relationships or utilizing the new terms.
Experimental - presents at least one experiment and its results.  Look for:  who did it, what are they trying to show, what did they measure, if there's more than one experiment, what are the differences, and if they make a mistake, what is it.  Questions usually involve applying the experimental information or doing calculations.
Persuasive Argument - presents explanations for a phenomena by more than one person.  Focus on the differences between arguments.  Questions are verbal-like (strenghten/weaken).

1. -When the s-orbital electron in Cr (in its electron configuration) is promoted to the d-orbital, is this called half-stability of subshell or orbital?
2. Stability of the subshell (s, p, d, and f are subshells).

2. (pg 58, Lesson Book): How did you know the passage is informational just from scanning? Why wouldn’t it be persuasive, since two separate models (quantum numbers vs. Bohr)?

Lots of new formulas --> informational.  And because it doesn't really contrast the models.

3. (pg 58, Lesson Book, Map): Don’t you need a descrtiption of each eqn, since it seems that the Eq1 Energy= Eqn3 Energy?

Why would you?  They both depend on the same things, so they work on the questions.

1. -Just from reading the passage initially, this equating of the two is not clear? Do we not care about this?

You're going too far with the passage.  Take it at face value.  If they ask you to compare them, look at their descriptive text (one is for Hydrogen only, the other is for any one-electron system).  Focus on relationships in the equations, not extraneous details.

4. (pg 57, Lesson Book, Map): When you state contrast keywords indicate where the author will change scope, what do you mean? Do you think of “scope” as the implicit portion of the paragraph, e.g. Kinematics for Airbag example as we saw in a previous class?

No, "scope" is what they're focusing on.  First, I'll focus on reaction time in safety, then I'll focus on seatbelts, then airbags.  This is a change in scope.

5. (pg 58, Lesson Book, Map Eq#3): If in Eq#3 you include the negative when you break the eqn into two proportionalities, then if a questions asks if you double Z and (1/3)n, wouldn’t the negative sign mistakenly be counted twice?

Nope.  The proportionality still holds.  In both cases, it's proportional to the NEGATIVE value.

6. (pg 58, Lesson Book): How short should paragraph maps be? What kind of keyword is “further” (located above Eq3)? Is it continuation is it more like the words furthermore or thus, i.e. a conclusion keyword?

Depends on how much new info is presented.  In general, short and sweet is how you want to go.  Further is continuation, but since they've set up three contrasting equations, we want to look at each of them.

7. (pg 58, Lesson Book): When you excite an electron, do you excite it into another shell (different n-value) or to a different subshell? When think of the notion that electrons at a lower principal quantum have lower energies do you rationalize it as electrons close to the positive nucleus make it lower in energy?

Could be to either a different shell or subshell.  And yes, think of it like gravitational potential energy.  The close it is to the middle (nucleus or Earth), the lower the energy.

8. (pg 58, Lesson Book): -When you go down a period Zeff↓, and the explanation that you had given had to do with electron-shielding effect. Wouldn’t the gain in protons and the subsequent gain of electrons neutralize this effect?

Valence electrons do NOT shield each other, so adding more protons increases the pull on each valence electron, and they can't block/shield for each other.

1. -How would you explain the following atomic radius trend: Cu>Ni>Cr? Also, why is the atomic radius of Zn > Cu?
2. Outside the scope of the MCAT.  They care about trends, not all the exceptions.  This is especially true for periodic trends.

9. (pg 64, Lesson Book, Critical Thinking): Why does sulfur have a greater release of energy (i.e. greater electron affinity) than both Si and P?

 Sulfur is the furthest to the right, so this follows our trend.  It is very happy picking up extra electrons because it "loves on" its electrons the most (greatest Zeff).  Thus, it has the greatest release of energy.

1. -In class, you had stated there are two places where you find positive EA values (i.e. energy needs to be added for addition of electron), These two places should be Group IIA and Group 8A (Noble Gases), correct?
2. Group IA and Noble Gases, not IIA.

10. (pg 66, Lesson Book): Why is HCl considered polar covalent as opposed to ionic?

These are both nonmetals, and nonmetals tend to make covalent bonds.  Also, the difference in electronegativity just isn't quite large enough to consider it ionic.

11. (pg 70, Lesson Book): What is the topic and scope of the passage?

Topic:  Chemical complexes
Scope:  Their chemistry, geometry, and general aspects

12. (pg 70, Lesson Book): When mapping, what do you ask yourself after reading each sentence to comprehend and make the most of the passage or sentence (i.e. to avoid glazing)?

Ask yourself, "do I need this?" and, if so, "what did this say?"

13. (pg 71, Lesson Book, #6): -How do you compare based on Charge and what would be the Total charge of each? If you do it as the passage does it (i.e. just look at the first ion), then the answer is different then correct answer?

Total charge, it says, is "of the species present (as absolute values)

Looking at the answers choices:

[Co(NH3)6]PO4 --> [Co(NH3)6]3+ + (PO4)3-.  Charges are +3 and -3.  As absolute values, |3|+|-3|=6
Na3[CoCl6] --> 3Na+ + [CoCl6]3-.  Charges are 3 atoms charged +1 and one charged -3.  As absolute values, 3|1|+|-3| = 6
[Cr(NH3)4Cl2]Cl --> [Cr(NH3)4Cl2]+ + Cl-.  Charges are +1 and -1.  As absolute values, |1|+|-1| = 2
[Cr(NH3)5Cl]SO4 --> [Cr(NH3)5Cl]2+ + (SO4)2-.  Charges are +2 and -2.  As absolute values, |2|+|-2| = 4 

1. -If you do it your way of Absolute Tot Charge, then answer choices A & B would be tied with +6charge. Plaease note that I understand your rationale for the # if ions approach.
2. Yes they would be tied.  But there's another criteria -- the number of ions.  And B definitely beats A for that.

14. (pg 73, Lesson Book, Charles’ Law, Graph of V vs. T that you drew on Board): -How would ↑Temp then ↑Volume, when a gas tries to occupy the volume of its container?

Assume a non-rigid container, like a balloon.  Increased temp will increase the volume of the balloon.

15. (pg 74, Lesson Book, 1^st Critical Thinking): -In class you said that an alternative way to answer the question would be is to see that if you double each of the number of moles of gas, then you get their partial pressures as well. Why and could you explain?

If there is a 1:1:1/2 ratio for the number of moles of gas, and they're all ideal gases, then we know that they'll pressures that also adhere to this ratio at the same temperature.  Thus, since we know that the total is 5 atm and because 2:2:1 is the same ratio as 1:1:1/2, then we can say that their pressures are 2 atm, 2 atm, and 1 atm respectively (that adds up to 5 atm!)

16. (pg 74, Lesson Book, 2^nd Critical Thinking): Is Grahm’s Law used for effusion or diffusion? IS the second critical thinking question an example of diffusion?

Both.  The second question is diffusion, yes.

17. (pg 76, Lesson Book, passage): What is the independent variable pressure or volume?

Neither is independent.  They both depend on the amount of mercury the student added.  It probably the most sense, though to have pressure be the X-axis, for purely conventional reasons.

18. (pg 77, Lesson Book, Q#9, Answer Choice II): Why would or how would you explain why CO₂ has stronger attractive forces than He (since CO₂ has not net dipole)?

It's bigger, and will therefore have more dispersion forces as well as more gravitational pull between molecules.

1. -In general, what would you do (or does it even happen) where the idea of am answer choice is correct (i.e. attractive forces) but the logic doesn’t apply for the specific case used?

Well, then it can't be the answer.

2. (Answer Choice I): usually pressure of a gas is affected by a volume of container and not gas (i.e not gas molecules pushing on other gas molecules)? So, if you take up more space wouldn’t you have greater pressure?
3. Pressure has nothing to do with molecules pushing on each other, as you said.  Thus, the size of the particles has nothing to do with measuring the pressure.