General Chemistry Question Thread

[QofQuimica]

All users may post questions about MCAT, DAT, OAT, or PCAT general chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what general chemistry topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
-general, MCAT-level gen chem.
-particular MCAT-level gen chem problems, whether your own or from study material
-what you need to know about gen chem for the MCAT
-how best to approach to MCAT gen chem passages
-how best to study MCAT gen chem
-how best to tackle the MCAT physical sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

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If you really know your gen chem, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the General Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university general chemistry TA teaching experience. In addition, I teach general chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 14 on PS, 43 overall.

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the PS section of the MCAT, and 36 overall.

-Sparky Man: Sparky Man has his Ph.D. in physical chemistry. He scored 14 on the PS section of the MCAT, and 36 overall.

-GCT: GCT scored in the 99th percentile on the PCAT. He has also taught introductory physics and general chemistry.

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[PalmettoGuy]

Howdy,
In my TPR book, it says that "When a system is undergoing a phase change, the heat added to that system adds to the potential energy of the molecules of that system", which makes since b/c during a phase change T=constant and KE=3/2RT therefore KE=constant.

However, in MCAT practice test 3R, a question is

"Evaporation occurs when molecules at the surface of a liquid overcome the attractive forces of the liquid. This occurs when molecules in the liquid attain a sufficient amount of:

B)Kinetic Energy
D)Potential Energy

I said D, the correct answer is B because:

"Evaporation occurs when a molecule attains sufficient speed or KE to overcome the attractive forces of a liquid. Resonance, surface tension and PE all relate to molecules that are not in motion. Thus, the answer choice B is the best answer."

Can anyone point out where I went wrong

Thanks

 

[free_radical]

i think when 3R says evaporation, they dont mean a phase change per se, otherwise you'd be correct that kinetic energy cant change, because temperature isnt changing.

i think they mean in general, when you have a pot of water, and molecules are jumping out of the surface, it must be an issue of KE. it helps to remember that in evaporation, the molecules are physically moving- ergo KE is an issue.

 

[BrokenGlass]

Howdy,
In my TPR book, it says that "When a system is undergoing a phase change, the heat added to that system adds to the potential energy of the molecules of that system", which makes since b/c during a phase change T=constant and KE=3/2RT therefore KE=constant.

However, in MCAT practice test 3R, a question is

"Evaporation occurs when molecules at the surface of a liquid overcome the attractive forces of the liquid. This occurs when molecules in the liquid attain a sufficient amount of:

B)Kinetic Energy
D)Potential Energy

I said D, the correct answer is B because:

"Evaporation occurs when a molecule attains sufficient speed or KE to overcome the attractive forces of a liquid. Resonance, surface tension and PE all relate to molecules that are not in motion. Thus, the answer choice B is the best answer."

Can anyone point out where I went wrong

Thanks

Vaporization is any change from a liquid to a gas (this includes evaporation and boiling). Vaporiazation is a state change, so the temperature remains constant during vaporization. Energy is needed to separate particles that attract each other. The energy input goes into breaking intermolecular forces. If you increase the distance between objects who extert attractive forces on each other, you increase their potentail energy.

Evaporation applies specifically when molecules at the SURFACE escape from a liquid BELOW the boiling point of that substance. Energy input is not required for evaporation. Remember that the AVERAGE kinetic energy depends on the temperature. While most of the molecules are near this average kinetic energy (temperature), there are a few molecules that are much more active (hotter) than the average. A few of these fast(hot) molecules can have enough energy to overcome the attractive forces (break away) from the neighboring molecules. When these faster molecules escape from the surface of the liquid you have evaporation. Since only these very high energy molecules can escape, it is not yet boiling.

Btw, escape velocity as it applies to escaping the gravitational pull of the earth is sort of similar to escape velocity as it applies to surface particles of the liquid breaking free from attractive forces exerted upon them by other liquid particles. In both cases, enough KE (i.e. enough velocity) is needed to "break free."

 

[AWhitehair]

Nevermind...I think I get it now.

Wouldn't osmotic pressure increase with increasing molarity of a liquid? I just got this q wrong... OP=MRT

The contractile vacuole of Euglena decreases its rate of contraction when the organism is transferred from fresh water to seawater. This is explained by

a) an increase in the osmotic pressure of the environment.
b) a decrease in the osmotic pressure of the environment.
c) the nitrogenous wastes can remain in the cell because of a higher salt concentration outside the cell.
d) excess salts are eliminated without the loss of water.
e) salt inhibits the contractile apparatus.

The correct answer is B???

 

[xlr8]

edit: was given the wrong answer to the question

 

[scotties123]

I couldnt find this online or in my exam krackers book. For galvanic cells and electrolytic cells, what is the equation for total cell potential. I know it has some form of: E = E(anode) +/- E(cathode)

And im pretty sure i remember it being different for galvanic versus electrolytic. Thanks.

 

[BrokenGlass]

I couldnt find this online or in my exam krackers book. For galvanic cells and electrolytic cells, what is the equation for total cell potential. I know it has some form of: E = E(anode) +/- E(cathode)

And im pretty sure i remember it being different for galvanic versus electrolytic. Thanks.

A galvanic cell has a positive emf. Since delta G = -nFE, E has to be positive for the reaction to be spontaneous. To find E for a galvanic cell, simply take the reduction potentials for the half reactions and add them up. But before you do that, flip the sign on one of the reduction potentials so that the resulting sum is positive.

 

[BrokenGlass]

The experiment in the passage was an emulation of Boyle's experiment, using He and adding increasing amounts of Hg while measuring the Pressure, Volume, and P*V.

If the student repeats the experiment at a higher temperature using xenon instead of helium, how would the results be affected?

A. Neither the large size of xenon nor the higher temperature would affect the results
B. The larger size of xenon would have no effect on the results, but the higher temperature would cause deviation from ideal behavior.
C. The larger size of xenon would result in deviation from ideal behavior, but the temperature change would have no effect on the results.
D. The larger size of xenon and the increased temperature would result in deviation from ideal behavior.

I chose C because of the larger size deviation, and also I remember reading that lower temperature is what causes deviation from ideal behavior because as the average KE of the molecules is slowed, the intermolecular attractions become more prominent.

The answer to the question is D. Does this mean that any type of temperature change causes deviation from ideal behavior of gases?

[SIZE=+1]Under what circumstances will the concept of the Ideal Gas and Kinetic Molecular Theory breakdown most significantly?[/SIZE]

1. Very large volume gas molecules will deviate significantly from Ideal behavior. Since the Ideal Gas is able to move anywhere in the container, if the particles are of large volume, then they will be restricted from parts of the container that are occupied by the other gas particles. Remember that the V variable is the volume of the container and also the volume available to the particles. If the two ideas differ significantly, then Ideal behavior falls apart. Conversely, the most Ideal gas based on volume will be a gas with a very small volume. Theoretically, Helium is the smallest volume gas commonly found in our environment. Helium is the most Ideal. Note that Hydrogen could be smaller except for the fact that it is most commonly found as a diatomic molecule.
2. Gas particles that exhibit nonpolar qualities are going to be more Ideal. Any gas that is highly polar, such as water, will experience significant attractions for the other particles in the system This will create problems with the concept of Ideal gases not interacting with each other. The Noble Gases, with their nonpolar character, will be the closest to Ideal in behavior.
3. Overall, for all gases, when the system is at high pressure or low temperature, the deviation from Ideal behavior will be substantial. These circumstances cause the particles to be close together and improve chances for interactions. Conversely, when gases are at low pressures and high temperatures, then they will be more Ideal. These circumstances will allow gases to move around with less attraction for one another and have less volume taken up on a percentage basis by the particles themselves.

 

[pezzang]

Why would increasing atm pressure favor reverse reaction below:
HI(aq) -> <- HI (g)
They are at the same mole so shouldn't the equilibrium be not disturbed?

 

[gridiron]

Why would increasing atm pressure favor reverse reaction below:
HI(aq) -> <- HI (g)
They are at the same mole so shouldn't the equilibrium be not disturbed?

Hey! Think of it in this way: when you increase the pressure what happens? The volume decreases. This means, according to the kinetic molecular theory, there will be a phase change because the particles will interact more and will stick together. Thus, the reverse reaction will be favored. Just because the moles are distributed equally on both sides doesn't necessarily mean equilibrium. You can only make that conclusion if both sides had the same phase. I hope this helps and good .

 

[cloosh]

this may have been brought up before but in regards to periodic table trends, would the test makers actually ask us a question of WHY certain trends happen, or is it just enough to know the trends?

 

[BirenJ]

I'm confused about critical temperature/pressure/point concept.

Critical temperature is the temperature, after which, the substance cannot be liquified regardless of pressure. So how would you read the critical temperature on a phase diagram? I'm totally lost because one of the EK questions has a critical temperature listed as 31. But on the phase diagram (temp vs pressure), after this temp, the substance can exist as gaseous or solid if you increase/decrease the pressure enough. Can anyone clarify this confusion?

 

[BrokenGlass]

this may have been brought up before but in regards to periodic table trends, would the test makers actually ask us a question of WHY certain trends happen, or is it just enough to know the trends?

http://forums.studentdoctor.net/showthread.php?t=210973

See posts 2 and 3.

 

[BrokenGlass]

I'm confused about critical temperature/pressure/point concept.

Critical temperature is the temperature, after which, the substance cannot be liquified regardless of pressure. So how would you read the critical temperature on a phase diagram? I'm totally lost because one of the EK questions has a critical temperature listed as 31. But on the phase diagram (temp vs pressure), after this temp, the substance can exist as gaseous or solid if you increase/decrease the pressure enough. Can anyone clarify this confusion?

http://www.chemguide.co.uk/physical/phaseeqia/phasediags.html

Scroll to the part that explains critical point.


Basically, at a temperature above the critical temperature, Tc, and a pressure above the critical pressure, Pc, it is no longer possible to distinguish between the gas and liquid phases. At T > Tc and P > Pc the substance is referred to as a super-critical fluid.

Sorry I cannot give you better answer because I don't understand your question well enough...

 

[BirenJ]

Alright, I see it now. The line seperating liquid and gas phases on the phase diagram disappears after the temperatures above the critical temperature.

EK's explanation of critical temperature just threw me off. Thank you.

 

[PalmettoGuy]

For two different electrochemistry problems, I have been asked if the density of an aqueous solution changes as sulfate ions react out of solution.

Pb(s) + PbO2(s) + 2 H2SO4(aq) -> 2 PbSO4(s) + 2 H20

So basically an aqueous soln of sulfate ions is decreasing in conc. No additional information is given on the densities.

NaCl increases the density of water (the ocean), and the SO4- ion also increases the density of water according to these passages. However, I would assume that a soluble organic molecule such as methanol or ethanol would decrease the density of an aqueous solution due to their nonpolar "tail" (correct me if I'm wrong). Any other examples should we know about? What should we know concerning how solutes/ions change the densities of solutions?

Thanks folks...

 

[wes431]

An electron in a certain element can have energies of -2.3, -5.1, -5.3, -8.2, and -14.9 eV. -14.9 eV is the ground state of the electron, and no other levls exist between -14.9 and -2.3 eV. Which of the following represents a partial list of photon energies that could be absorbed by an electron in the ground state of the electron? All energies are in electron volts.

A. -2.3, -5.1, -5.3, -8.2, -14.9
B. 0.2, 2.8, 2.9, 6.7
C. 2.3, 5.1, 5.3, 8.2, 14.9
D. 6.7, 9.6, 9,8, 12.6, 15.0, 16.1

Why is D the right answer?

 

[jochi1543]

An electron in a certain element can have energies of -2.3, -5.1, -5.3, -8.2, and -14.9 eV. -14.9 eV is the ground state of the electron, and no other levls exist between -14.9 and -2.3 eV. Which of the following represents a partial list of photon energies that could be absorbed by an electron in the ground state of the electron? All energies are in electron volts.

A. -2.3, -5.1, -5.3, -8.2, -14.9
B. 0.2, 2.8, 2.9, 6.7
C. 2.3, 5.1, 5.3, 8.2, 14.9
D. 6.7, 9.6, 9,8, 12.6, 15.0, 16.1

Why is D the right answer?

Ground state is -14.9 and we have a list of all the other states that the electron can jump to. For an electron to jump from ground state to any other state, it must absorb a photon of the corresponding energy difference. If you take a close look, the difference between -8.2 and -14.9 is 6.7, which is the first # in D. Between -5.3 and 14.9 it's 9.6, which is the 2nd #...you get the drift.

 

[wes431]

Ground state is -14.9 and we have a list of all the other states that the electron can jump to. For an electron to jump from ground state to any other state, it must absorb a photon of the corresponding energy difference. If you take a close look, the difference between -8.2 and -14.9 is 6.7, which is the first # in D. Between -5.3 and 14.9 it's 9.6, which is the 2nd #...you get the drift.

Alright I get that, but what about the 15.0 and 16.1 in D? Where do they come from?

 

[jochi1543]

Alright I get that, but what about the 15.0 and 16.1 in D? Where do they come from?

Haha, I don't know, I'm of the "don't spend too much time thinking" school of thought (pun intended). On the MCAT, there'll be a ton of options you won't understand, but you have to just ignore the things that are beyond your comprehension and go by what you know.

 

[wes431]

lol alright. also, what does it mean, when they say energy is quantized?

 

[wes431]

Also another question. I was doing a problem and one of the answer choices was "the bonds are stronger because larger atoms are more polarizable as period increases." This ended up being the right answer, but I have no clue what polarizable means. Anyone know?

 

[EmiliaC]

lol alright. also, what does it mean, when they say energy is quantized?

My guess is that "quantized" refers to energy being released in discrete bundles, i.e. an electron moving to a lower orbital from a higher one releases a photon of a specific wavelength with a specific amount of energy.

Also another question. I was doing a problem and one of the answer choices was "the bonds are stronger because larger atoms are more polarizable as period increases." This ended up being the right answer, but I have no clue what polarizable means. Anyone know?

I think "polarizable" means that the atom is able to become (at least temporarily) polarized, with + and - charges distinct from each other enough to influence bonding properties. This might occur in a larger atom with electrons orbiting far from the nucleus when several outer shell electrons move near each other, creating a temporary aggregation of negative charges while the nucleus remains positively-charged (dipole).

 

[BrokenGlass]

Alright I get that, but what about the 15.0 and 16.1 in D? Where do they come from?

There aren't enough energy levels provided to deduce the 15.0 and 16.1 transitions directly from math, but clearly answer choice D is the best answer, so the assumption here is IF more energy levels were given, 15.0 and 16.1 could have been deduced by doing math.

 

[ashley]

As I recall learning in class, the conjugate base of a strong acid was described as a "weak base" and the conjugate base of a weak acid was a "strong base." (similar principle applied to the conjugate acids of strong/weak bases)

However, going over my TPR material, it says "the conjugate base of a weak acid is a weak base" and that "the conjugate base of a strong acid has no basic properties in water."

Am I then to understand that the definitions given in school were merely relative?

 

[pezzang]

In TPR test, it says "According to Le Chantelier's principle, an increase in T favors an endothermic rxn, so if a solvation process is endothermic, the solubility will increase with increasing T. For exothermic solvations, the reverse is true."


1. I always thought that increasing T increases solubility of solid in liquid no matter whether the solvation process is endo/exothermic. So shouldn't increasing T increase solubility in EXOthermic solvation as well (the TPR statement "the reverse is true" contradicts this fact).

2. I realize Le Chantelier;s principle works under the assumption that the process is in equilibrium, so while something is solvating, the rate of forward rxn is NOT equal to the rate of reverse rxn... otherwise, it would be fully satured (equilibrium in which forward rxn (solvating) is equal to reverse rxn (precipitating). So how is it possible to use Le Chantelier's principle to test rxn process in non-equilibrium process as shown by the TPR statement above?

Sorry for the long post but this statement really bothers me...

 

[BrokenGlass]

As I recall learning in class, the conjugate base of a strong acid was described as a "weak base" and the conjugate base of a weak acid was a "strong base." (similar principle applied to the conjugate acids of strong/weak bases)

However, going over my TPR material, it says "the conjugate base of a weak acid is a weak base" and that "the conjugate base of a strong acid has no basic properties in water."

Am I then to understand that the definitions given in school were merely relative?

pKa + pKb = 14 (for a conjugate pair at 25 degrees Celsius)

pka -----strong acid------0------weak acid-----------14----very weak acid----------

pkb -----strong base-----0-------weak base----------14------very weak base--------

So a weak base has a pKb between 0 and 14. If you subtract a number between 0 and 14 from 14, you
get a number between 0 and 14, which describes pKa for a weak acid.

So conj. base of a strong acid is a very weak base, conj. base of a very weak acid is a strong base,
conj. base of of weak acid is a weak base.

Conj. acid of a strong base is a very weak acid, conj. acid of a very weak base is a strong acid,
conj. acid of a weak base is a weak acid.

In other words, weak pairs up with weak, strong pairs up with very weak.

I think this level of understanding should be enough for the MCAT.

 

[BrokenGlass]

In TPR test, it says "According to Le Chantelier's principle, an increase in T favors an endothermic rxn, so if a solvation process is endothermic, the solubility will increase with increasing T. For exothermic solvations, the reverse is true."


1. I always thought that increasing T increases solubility of solid in liquid no matter whether the solvation process is endo/exothermic. So shouldn't increasing T increase solubility in EXOthermic solvation as well (the TPR statement "the reverse is true" contradicts this fact).

2. I realize Le Chantelier;s principle works under the assumption that the process is in equilibrium, so while something is solvating, the rate of forward rxn is NOT equal to the rate of reverse rxn... otherwise, it would be fully satured (equilibrium in which forward rxn (solvating) is equal to reverse rxn (precipitating). So how is it possible to use Le Chantelier's principle to test rxn process in non-equilibrium process as shown by the TPR statement above?

Sorry for the long post but this statement really bothers me...

Most aqueous solutions involving liquid or solid solutes will have ENDOTHERMIC heats of solution. In other words, the heat given off in the dissolving reaction is LESS than the heat required to break apart the solid.

Solute + Solvent + Heat ----> Solution (endothermic reaction)

In such cases, an increase in temperature produces an increase in solubility for solids, i.e. the reaction shifts in the endothermic direction, as predicted by La Chatalier's principle.

--------------------------------------------------------------------------------------------------------------------------

Other solutions will release thermal energy during the solution formation process. Put another way, the heat given off in the dissolving process is GREATER than the heat required to break apart the solid. These solutions are said to have an EXOTHERMIC heat of solution, e.g. Ce2(SO4)3.

Solute + Solvent ----> Solution + Heat (exothermic reaction)

In such cases, an increase in temperature produces a decrease in solubility, i.e. the reaction shifts in endothermic direction, as predicted by La Chatalier's principle.

------------------------------------------------------------------------------------------------------------------------

Generally speaking the solubility of a liquid or solid will increase with increasing temperature (but again, there are some exceptions to this). Gaseous solutes always have an exothermic heat of solution. Consequently, the solubility of gases decrease with increasing temperature.


Assuming I understand your second question correctly, when a solution mixture reaches equilibrium, and we add apply stress to it in the form of added heat, the equilibrim will shift to alleviate the stress and it will shift in the endothermic direction.

Post back if you have any more questions....

 

[Creightonite]

Hess's law question.

This applied for calculating the deltaG, H and Hf for the reaction. What I do not understand is that there are two basic equations:

1) deltaH=H(products) + H(reactants)

or

2) deltaH=H(products) - H(reactants)

They would usually give several reaction and you just to have invert the correct reaction and extract the numbers from that. When are we supposed to use equation #1 or #2. Kaplan book gives them both but no real explanation in the difference in their use. Is the difference comes when you calculate deltaG or deltaH or Hf? or elsewhere?

Thank you.

 

[cheezer]

Hess's law question.

This applied for calculating the deltaG, H and Hf for the reaction. What I do not understand is that there are two basic equations:

1) deltaH=H(products) + H(reactants)

or

2) deltaH=H(products) - H(reactants)

They would usually give several reaction and you just to have invert the correct reaction and extract the numbers from that. When are we supposed to use equation #1 or #2. Kaplan book gives them both but no real explanation in the difference in their use. Is the difference comes when you calculate deltaG or deltaH or Hf? or elsewhere?

Thank you.

I'm just learning this in class, but you use them when you don't have several reactions, just the main one. then you just look up heat of formation in a chart somewhere and plug them into the equation, taking into account the moles. i was taught equation number two, don't know about one, can someone elaborate?

 

[BrokenGlass]

Hess's law question.

This applied for calculating the deltaG, H and Hf for the reaction. What I do not understand is that there are two basic equations:

1) deltaH=H(products) + H(reactants)

or

2) deltaH=H(products) - H(reactants)

They would usually give several reaction and you just to have invert the correct reaction and extract the numbers from that. When are we supposed to use equation #1 or #2. Kaplan book gives them both but no real explanation in the difference in their use. Is the difference comes when you calculate deltaG or deltaH or Hf? or elsewhere?

Thank you.

METHOD 1:

The correct formula is deltaH=deltaH(products) - deltaH(reactants). Also, don't forget the stoichiometric coefficients in this formula.
So make sure the reaction is correctly balanced before you use this formula.

METHOD 2:

Several reactions may need to be manipulated ot obtain the overall reaction. Some of the reactions may need to be reversed so that the products and the reactants end up on the correct side (with respect to the overall reaction). When you reverse a reaction, you reverse the sign of delta H. When you mulitply a reaction by some constant, you mulitply the delta H by this same constant.

Again, the goal is to manipulate a set of reactions such that their sum adds up to the overall reaction. While you are doing this, you need to make the corresponding adjustments to the delta H values.

HTH

 

[Creightonite]

EDIT: NO ANSWER NEEDED.... I now KNOW THE ANSWER ~~~~~~

Reduction potential question.

Kaplan book says that higher E --> the element more likely to be reduced, while lower E--> the element likely to be oxidized

so they give two reactions:

Ag+ and e --> Ag(s) E=+0.80 V
Tl+ and e --> Tl(s) E=-0.34V

Since Tl's voltage is lower then it will be oxidized and the following reaction will look like this:

Ag+ and Tl (s) --> Tl+ and Ag(S)

My question: notice that Tl+ equation gets inverted. So in order to decide whther something gets oxidized or reduced, should we then invert Tl+ equation to make it into an oxidation (making it a +0.34V) and then compare voltages? Or it is a strict reduction potential comparison? Because if we want to get the total E for thr whole reaction we would have to do this:

E= +0.80 + 0.34 = 1.14V.

Is my thinking correct?

 

[Creightonite]

one more question (comes from Kaplan FL#2).

is the equavalence point of adding weak base to strong base going to be higher or lower than when addition strong acid to strong base

Answer: higher

I really do not understand their explanation. Can someone explain it? All I can think that the weak base wont be able to resist the pH change that much when being titrated with a SA and it gives up a little raising the equivalence point. I cant think of it in scientific terms though... any help?

 

[Hindiana_Jones]

Okay, from my understanding, exothermic is when heat is released by the system. The &#916;H is negative. I also thought that when bonds form, the process is exothermic. Endothermic reactions have a positive &#916;H, and when bonds break the process is endothermic because energy is required to break the bonds. Ice melting is endothermic, the ice absorbing the heat from the surrounding, etc, etc.

So for this problem, they tell me the standard enthalpy of formation for liquid H2O is is -285.8 kJ/mol, which makes sense in my mind because going from a gas to a liquid would require forming bonds, which is an exothermic process and would release energy. The question asks me what the standard enthalpy of formation for water vapor?

So I thought this, to go from liquid water to water vapor, bonds are breaking. Also it would require a high energy input to break the bonds. So I thought this process would be endothermic and have a positive &#916;H. But this was wrong.

These were the options:
A. -480.7 kJ/mol
B. -285.8 kJ
C. -241.8 kJ/mol
D. +224.6 kJ/mol

So I picked D. The correct answer is C, and Ek says the answer is obviously negative, and that the condensation of water vapor should be more exothermic than vaporization. So, what is wrong with my line of thinking? How have I screwed up the concepts of exothermic and endothermic?
Is it because &#916;H rxn = total energy input - total energy output, and so the input to break the bonds is less than the energy released, making the overall reaction exothermic?

 

[clc8503]

I'm having trouble with this particular concept of General Chemistry. I will appreciate any help I can get.

Thanks!

Consider the following gas molecules: CH4 CF2Cl2 CHCl3 CH3Cl CH2F2

(a) At 25.0 °C, which would be moving the fastest?


(b) At 25.0 °C, which molecule would have the most energy?


(c) If the above gas molecules were all moving at he same speed, put them in order of increasing temperature.

< < < <


Thank you so much!

 

[gridiron]

Okay, from my understanding, exothermic is when heat is released by the system. The &#916;H is negative. I also thought that when bonds form, the process is exothermic. Endothermic reactions have a positive &#916;H, and when bonds break the process is endothermic because energy is required to break the bonds. Ice melting is endothermic, the ice absorbing the heat from the surrounding, etc, etc.

So for this problem, they tell me the standard enthalpy of formation for liquid H2O is is -285.8 kJ/mol, which makes sense in my mind because going from a gas to a liquid would require forming bonds, which is an exothermic process and would release energy. The question asks me what the standard enthalpy of formation for water vapor?

So I thought this, to go from liquid water to water vapor, bonds are breaking. Also it would require a high energy input to break the bonds. So I thought this process would be endothermic and have a positive &#916;H. But this was wrong.

These were the options:
A. -480.7 kJ/mol
B. -285.8 kJ
C. -241.8 kJ/mol
D. +224.6 kJ/mol

So I picked D. The correct answer is C, and Ek says the answer is obviously negative, and that the condensation of water vapor should be more exothermic than vaporization. So, what is wrong with my line of thinking? How have I screwed up the concepts of exothermic and endothermic?
Is it because &#916;H rxn = total energy input - total energy output, and so the input to break the bonds is less than the energy released, making the overall reaction exothermic?

There was a discussion in the MCAT forum not too long ago. See if this helps:

http://drslounge.studentdoctor.net/showthread.php?p=5320068

 

[gridiron]

I'm having trouble with this particular concept of General Chemistry. I will appreciate any help I can get.

Thanks!

Consider the following gas molecules: CH4 CF2Cl2 CHCl3 CH3Cl CH2F2

(a) At 25.0 °C, which would be moving the fastest?


(b) At 25.0 °C, which molecule would have the most energy?


(c) If the above gas molecules were all moving at he same speed, put them in order of increasing temperature.

< < < <


Thank you so much!

Hey!

(a) room mean square velocity is dependent on molecular mass and temperature. For this case, the temperature is the same for each molecule. The heavier the molecule, the slower it will move. The fastest molecule will have the smallest molecular mass. The equation for rms velocity is:

v rms = square root (3RT/M)

where M is the molecular mass

(b) If all the molecules are at the same temperature, they will all have the same energy. This is a TRICK question and a good MCAT question. Kinetic energy for a molecule is given by: 3/2kT. Since all molecules are at the same temperature, they will all have the same energy.

(c) This question seems to be the opposite of (a). If you solve for the temperature in the equation for (a), temperature will be directly proportional to molecular. That means the heavier molecule will have the highest temperature---someone correct me if I am wrong.

Hope this helps and good .

 

[clc8503]

Hey!

(a) room mean square velocity is dependent on molecular mass and temperature. For this case, the temperature is the same for each molecule. The heavier the molecule, the slower it will move. The fastest molecule will have the smallest molecular mass. The equation for rms velocity is:

v rms = square root (3RT/M)

where M is the molecular mass

(b) If all the molecules are at the same temperature, they will all have the same energy. This is a TRICK question and a good MCAT question. Kinetic energy for a molecule is given by: 3/2kT. Since all molecules are at the same temperature, they will all have the same energy.

(c) This question seems to be the opposite of (a). If you solve for the temperature in the equation for (a), temperature will be directly proportional to molecular. That means the heavier molecule will have the highest temperature---someone correct me if I am wrong.

Hope this helps and good .

Thanks!

How sure are you about (b)? It was the main one that I was confused on. I did some reserch and I completely agree with you on (a) and (c). I'm not saying (b) is wrong. Like you said, it is a trick question...

 

[axp107]

Solubility problems =(

For some reason, whenever I see a solubility problem I just freeze.. I can't think.. my brain just stops working

For example...

1M NaOH is added to a solution containing 1 M Ag+, Al3+, Mg2+, and 1 M Mn2+. Given the solubility data below, which will precipite first?

Ksp values:

AgOH = 1.5E-8
Al(OH)3 = 3.7 E -15
Mg(OH)2 = 1.2 E -11
Mn(OH)2 = 2E-13


How do I even START this problem?

I call solubility x

I set up equations for the rxns... and looked for the LOWEST solubility, meaning lowest x

I got...

AgOH -> Ag+ + OH-
We know Ag+ is 1... so

ksp AgOH = 1.5E-8

so

1.5E-8 = x

x is the lowest value for AgOH compared to all the other compounds (i tried it)

So when asked what precipitates first, are we looking for the lowest solubility, x?

 

[BrokenGlass]

Thanks!

How sure are you about (b)? It was the main one that I was confused on. I did some reserch and I completely agree with you on (a) and (c). I'm not saying (b) is wrong. Like you said, it is a trick question...

http://forums.studentdoctor.net/showthread.php?t=418733

 

[BrokenGlass]

EDIT: NO ANSWER NEEDED.... I now KNOW THE ANSWER ~~~~~~

Reduction potential question.

Kaplan book says that higher E --> the element more likely to be reduced, while lower E--> the element likely to be oxidized

so they give two reactions:

Ag+ and e --> Ag(s) E=+0.80 V
Tl+ and e --> Tl(s) E=-0.34V

Since Tl's voltage is lower then it will be oxidized and the following reaction will look like this:

Ag+ and Tl (s) --> Tl+ and Ag(S)

My question: notice that Tl+ equation gets inverted. So in order to decide whther something gets oxidized or reduced, should we then invert Tl+ equation to make it into an oxidation (making it a +0.34V) and then compare voltages? Or it is a strict reduction potential comparison? Because if we want to get the total E for thr whole reaction we would have to do this:

E= +0.80 + 0.34 = 1.14V.

Is my thinking correct?

Reduction potentials are measured with respect to reduction potential for a proton. Reduction potential of proton is arbitrarily assigned a value of zero. Remember that reduction potential is just ENERGY per unit charge.

Now, if (standard) reduction potential is greater than zero, then the element is more eager to be reduced as compared to a proton.
If (standard) reduction potential is less than zero, then the element is less eager to be reduced as compared to a proton.

So if you have a pair of reduction half-reactions and you want to know which reaction you need to reverse in order to convert it into an oxidation reaction, you need to look at both reactions and aks yourself which of the 2 likes to be reduced less. That's the reaction that you need to convert to an oxidation reaction.

Another way is to remember that in a galvanic cell, the cell potential is always positive. (delta G = -nFE, so in order to get a favorable overall reaction, E has to be positive). So you need to reverse one of the half reactions such that the sum of potentials is positive.

 

[BrokenGlass]

one more question (comes from Kaplan FL#2).

is the equavalence point of adding weak base to strong base going to be higher or lower than when addition strong acid to strong base

Answer: higher

I really do not understand their explanation. Can someone explain it? All I can think that the weak base wont be able to resist the pH change that much when being titrated with a SA and it gives up a little raising the equivalence point. I cant think of it in scientific terms though... any help?

First, remember that in ALL titrations, the titrant (the substance that you are adding) is STRONG, while the analyte (the substance that's already there) may be strong or weak.

When you add a strong base to a strong acid, the pH at equivalence point is 7 (at 25% Celsius). Why? Because the products are neutral salt and water. So [H+] = [OH-].

When you add a strong base to a weak acid, the pH > 7. Why? Because the strong base converts the weak acid into the conjugate base of the weak acid, so the product is a weak base.

When you want to know the pH at equivalence point, ask yourself what the products are.

 

[BrokenGlass]

Solubility problems =(

For some reason, whenever I see a solubility problem I just freeze.. I can't think.. my brain just stops working

For example...

1M NaOH is added to a solution containing 1 M Ag+, Al3+, Mg2+, and 1 M Mn2+. Given the solubility data below, which will precipite first?

Ksp values:

AgOH = 1.5E-8
Al(OH)3 = 3.7 E -15
Mg(OH)2 = 1.2 E -11
Mn(OH)2 = 2E-13


How do I even START this problem?

I call solubility x

I set up equations for the rxns... and looked for the LOWEST solubility, meaning lowest x

I got...

AgOH -> Ag+ + OH-
We know Ag+ is 1... so

ksp AgOH = 1.5E-8

so

1.5E-8 = x

x is the lowest value for AgOH compared to all the other compounds (i tried it)

So when asked what precipitates first, are we looking for the lowest solubility, x?

In solubility problems, you need to compare molar solubilities (the x values) and not solubility products (Ksp values) because molar solubilities always have the same units, while Ksp values have the same units only if we are dealing with the same types of salts.

For instance, Mg(OH)2 and Mn(OH)2 are the same types of salt.

For the problem you posted, we need to figure out the molar solubilities and then look for the LOWEST molar solubility (because the compound that likes to be dissolved least will precipitate first).

AgOH = Ag+ + OH-
x x x

So one mole of AgOH produces one mole of Ag+ and one mole of OH-. Ksp = x^2. Remember that there is no denominator for Ksp since the reactant is always a pure solid. (We never include pure solids and pure liquids into K. For this same reason, Kw, the autoionization constant for water has no denominator because water, a pure liquid, is the only reactant).

x^2 = 1.5E-8 => x = sqrt(1.5)*E-4

Al(OH)3 = 3OH- + Al3+
x 3x x

Ksp = (3x)^3*x = 27x^4
x = (Ksp/27)^1/4

Similary, for both Mg(OH)2 and Mn(OH)2, x= (Ksp/9)^1/3.

You do the math!!!!!!!!!!

Again, remember that it only makes sense to compare quantities whose units are the same. So the last 2 salts are the same types of salt, so we could eliminate Mg(OH)2 as the correct answer because Mn(OH)2 has a lower Ksp, so it will precipitate out before Mg(OH)2. But in general, you need to work with molar solubilities and not solubility products.

Also, after you do the math and start comparing your x values, just focus on powers of 10 to see which value is the smallest.

One last comment: to take a square root, it's easiest if you first convert opearand into scientific notation where the exponent is a multiple of 2. This way you could simply divide the exponent by 2. To to take a cube root, it's easiest if you first convert opearand into scientific notation where the exponent is a multiple of 3. This way you could simply divide the exponent by 3.

HTH

 

[xlr8]

"conjugated double bonds systems stabilize the electrons, so that they can be excited by lower-frequency photons with wavelengths in the visible spectrum."

If the electrons are more stable, wouldn't a higher energy photon (and thus a higher frequency photon) be needed to excite the electron?

 

[BrokenGlass]

"conjugated double bonds systems stabilize the electrons, so that they can be excited by lower-frequency photons with wavelengths in the visible spectrum."

If the electrons are more stable, wouldn't a higher energy photon (and thus a higher frequency photon) be needed to excite the electron?

For electron transitions what matters is the RELATIVE distance between energy levels. So if both energy levels get lowered, such that n+1 energy level gets lowered more than the n energy level, the photon that causes n->n+1 transition would be less energetic, meaning lower frequency (E = h*f), higher wavelength (c = f*lambda, lambda = c/f) and visible spectrum (apparently).

I am just saying that there is nothing wrong with the sample solution. But since I don't have the complete problems statement and the answer key, I cannot give you a better answer.

 

[xlr8]

For electron transitions what matters is the RELATIVE distance between energy levels. So if both energy levels get lowered, such that n+1 energy level gets lowered more than the n energy level, the photon that causes n->n+1 transition would be less energetic, meaning lower frequency (E = h*f), higher wavelength (c = f*lambda, lambda = c/f) and visible spectrum (apparently).

I am just saying that there is nothing wrong with the sample solution. But since I don't have the complete problems statement and the answer key, I cannot give you a better answer.

I was under the impression that more stability equates to a higher energy (and frequency) of photons needed to excite an electron

So it is possible that for instance an n=4 energy level becomes more stable than an n=3 energy level from the conjugation, making the 3 -> 4 transition require less energy?

 

[BrokenGlass]

I was under the impression that more stability equates to a higher energy (and frequency) of photons needed to excite an electron

So it is possible that for instance an n=4 energy level becomes more stable than an n=3 energy level from the conjugation, making the 3 -> 4 transition require less energy?

No, an n+1 level is always higher in energy than an n level. When you think of an electronic transition, think about "where from" and "where to." It doesn't matter where any given level is. What matters in transition is the relative energy levels between starting and ending energy states.

 

[xlr8]

Is the original statement I posted saying that the stabilization allows for more excitations to occur, specifically ones with lower energy differences, and lower frequencies?

 

[BrokenGlass]

Is the original statement I posted saying that the stabilization allows for more excitations to occur, specifically ones with lower energy differences, and lower frequencies?

The original statement you posted was not very clear. I tried to take my best shot at it, but unless you post the entire problem statement and the answer key, it's hard to make meaningful comments.

 

[mpatricksweeney]

On this same topic...

The energy of an e is E = -Rydberg / n^2. According to this eqtn, e gains energy as it is separated from nucleus, ie energy value becomes less negative, but the magnitude of the energy is also reduced. Once ionized (n=infinity) the e's energy approaches 0.

I'm confused by the negative energy value. Could someone explain? For instance, when e moves further out (from n=1 to n=4) it is excited and is said to have gained energy (or lost negative energy). But as e moves from n=4 to n=10 (if 10 were possible), its energy is approaches zero-- ie has virtually NO energy. How can it have gained energy-- excitement-- and end up with no energy??

Thanks...