General Chemistry Question Thread 2

[gridiron]

This is a continuation of the original General Chemistry Thread.

---

Members don't see this ad.

 

[iA-MD2013]

Pretty sure that if you increase the pressure your reaction shifts to the side with fewer moles of gas, that is it would move to the product (2 mol NH3 vs. 4 mol H2 + N2). Assuming a raise in volume constitutes a reduction in pressure, I would think the opposite would occur. Your reaction would move to the right side.

What about adding an inert gas? TPR says that it has no effect...but isn't adding another gas the same as increasing the pressure/decreasing the volume?

 

[SketchLazy]

What about adding an inert gas? TPR says that it has no effect...but isn't adding another gas the same as increasing the pressure/decreasing the volume?

When you add an inert gas, you increase the total pressure, but you don't increase the partial pressures of each gas in the reaction. So in terms of Le Chatelier's Principle, it is like nothing happened.

 

[DrMattOglesby]

so if i had an equal number of moles on both sides, then increasing pressure or decreasing the volume wouldnt shift the reaction. at least that is my understanding

i was looking for someone to give a breakdown of the gas law: PV=nRT
where we see the inverse relationship of P=(nRT)/V

from this setup of the equation, i can see how an increase in pressure pushes the system to the address some situation with the moles...but i dont know how to understand or communicate that idea.

after some nose-picking, this is what i want to say--tell me if it makes sense to your brain:
Since the Pressure and moles are directly proportional, if you increase one, you increase the other.
Therefore, when pressure is increased, the system will be pushed in a way to produce an increase in moles as well...by decreasing the volume.

But here is the catch;
since PV=nRT can be rewritten as V=(nRT)/P, then an increase in volume will also favor an increase in moles, so the system would similarly shift to produce more moles?

 

[iA-MD2013]

When you add an inert gas, you increase the total pressure, but you don't increase the partial pressures of each gas in the reaction. So in terms of Le Chatelier's Principle, it is like nothing happened.

If you add enough of the inert gas, aren't you decreasing the volume?

 

[tncekm]

I'm sure this answer would be yes in real conditions, but it probably wouldn't apply to the ideal gas law where we assume that the gas molecules occupy zero volume.

 

[iA-MD2013]

I'm sure this answer would be yes in real conditions, but it probably wouldn't apply to the ideal gas law where we assume that the gas molecules occupy zero volume.

ohhhh i see...that makes sense. thank you!

 

[Kaustikos]

If you don't have the solubility constant or the concentration of MgCl2, then you wouldn't be able to figure out a trend. But let's just say you are given the Ksp's of both Barium hydroxide and magnesium hyrdoxide. The Ksp for magnesium hyroxide is lower than the Ksp for Barium hydroxide. So if there is any hydroxide present in solution, it will react with the magnesium ions and precipitate as Mg(OH)2 when the MgCl2 is added. The key to understand here is that magnesium hydroxide will form once you add MgCl2 even if magnesium hydroxide wasn't in the solution to begin with.

It's the same idea as when you add acid to the solution. The reason H+ reacts so readily with OH- is because Kw is 10^14. Since the K is so high, you're more likely to have water in solution than H+ and OH-. For magnesium hydroxide, the same idea applies. The K for it's formation is the inverse of the Ksp, ~6*10^10, meaning there will mostly be magnesium hydroxide rather than Mg+2 and OH- ions in solution. Just like how the H+ reacts with the OH- to form H20 shifting the equation to the right, the Magnesium reacts with the hydroxide to form magnesium hydroxide. Either way the hydroxide is sequestered because of a high equilibrium constant favoring either H2O or Mg(OH)2.

You know what? I know what I did there. I completely misunderstood the question he was asking (and still confused by the wording). So you're saying how MgCl would affect BaOH2 disassociation, which I understand how you came with your answer, and what TNCEKM meant. Sorry.

One last question; what the hell is the question actually asking?

 

[Kaustikos]

so if i had an equal number of moles on both sides, then increasing pressure or decreasing the volume wouldnt shift the reaction. at least that is my understanding

i was looking for someone to give a breakdown of the gas law: PV=nRT
where we see the inverse relationship of P=(nRT)/V

from this setup of the equation, i can see how an increase in pressure pushes the system to the address some situation with the moles...but i dont know how to understand or communicate that idea.

after some nose-picking, this is what i want to say--tell me if it makes sense to your brain:
Since the Pressure and moles are directly proportional, if you increase one, you increase the other.
Therefore, when pressure is increased, the system will be pushed in a way to produce an increase in moles as well...by decreasing the volume.

But here is the catch;
since PV=nRT can be rewritten as V=(nRT)/P, then an increase in volume will also favor an increase in moles, so the system would similarly shift to produce more moles?

I always looked at pressure as a measure of how concentrated your gas is (moles/volume). High pressure = high moles/volume. Low pressure = low moles/volume.

You have the right idea, though. When pressure increases, you're away from equilibrium and the system will do its best to shift back into equilibrium. It can do that by either changing volume or moles. Since we're limited to moles, it'll either reverse reaction rates or speed up formation of products.

Except for your first concept. Increased pressure will cause the system to favor the formation of lowest moles/partial pressures.


3X(g) + 2Y (g) <--> 2XY(g). Increase pressure and it'll go more towards the right, which favors the "lowering" of "concentration".

 

[bluemonkey]

When you add an inert gas, you increase the total pressure, but you don't increase the partial pressures of each gas in the reaction. So in terms of Le Chatelier's Principle, it is like nothing happened.

This is not entirely correct. As you said, it is an inert gas, so it will not react with the system. The addition of an inert gas leaves the partial pressures of the reacting species unchanged (the molar concentrations of the original gases are unchanged, because they continue to occupy the same volume). Thus, this type of pressurization has no effect on the equilibrium composition of the system (assuming the gases are perfect). This is only the case if the system is at a fixed volume. If, however, the system's volume is allowed to change, ie increase upon addition of the inert gas, then the partial pressures of all gases would decrease and the system will shift to produce more moles of gas in order to return to Kp.

 

[DrMattOglesby]

thanks man...this is surprisingly starting to make a little sense now.

 

[Kaustikos]

Need some more help with general chemistry.

First topic; solubility in terms of Ksp and Ion Product.

From my understanding, when you're trying to calculate the Ksp, you're using equilibrium concentrations and you multiply the concentration (solubility) by its exponent as well as taking it to the power:
XY2 <----> X + 2Y would mean
Ksp = [X][2 x X]^2 or 4X^3

But if you're going about it for the ion product, you're referring to initial concentrations and you DO NOT multiply but instead just use powers

XY2 <---> X + 2Y would mean
IP = [X][Y]^2.
Can someone tell me if I am correct and WHY this is true? I think I can see the answer/reasoning, but I am not sure. It seems that this has to do with the fact that the initial concentrations are not necessarily equal and so you cannot just say that there are twice as much Y as X like with Ksp. But I don't know. Kaplan is **** for explanations and expects you to figure it out. Am I right in saying that's how they differ? and why is that so?

Now here's the other dilemma: Entropy

E = q/T, so to speak. Entropy increases with disorder and increased temperature means increased kinetic energy which means more disorder. But now I am a little confused. WHY is the T variable at the bottom then? This would imply a higher T would mean a lower entropy value. But you would expect a higher entropy/disorder for higher temperature situations in a system, right? Or is this something dealing with the universe only?
Also, q is heat, which I am lead to think that a higher q means more heat added to the system, which means increased entropy, according to this equation. Is this also right?

Now the bigger dilemma: q added is endothermic for enthalpy, which is non-spontaneous. For entropy, q added is a plus, which makes G more negative and increases the sponteneouity of the reaction. Does this mean that heat added is something that requires one to factor the affects on enthalpy AND entropy?

Also, is it just me or is temperature neglible in the overall gibbs free energy reaction:
G = H - TS (S = q/T), so G = H - Q?
So gibbs free energy is more spontaneous with more heat added to the system, despite enthalpy thinking otherwise? Is enthalpy, therefore, the stubborn friend who just doesn't want to do X even though everyone else does?

 

[SketchLazy]

Need some more help with general chemistry.

First topic; solubility in terms of Ksp and Ion Product.

From my understanding, when you're trying to calculate the Ksp, you're using equilibrium concentrations and you multiply the concentration (solubility) by its exponent as well as taking it to the power:
XY2 <----> X + 2Y would mean
Ksp = [X][2 x X]^2 or 4X^3

But if you're going about it for the ion product, you're referring to initial concentrations and you DO NOT multiply but instead just use powers

XY2 <---> X + 2Y would mean
IP = [X][Y]^2.
Can someone tell me if I am correct and WHY this is true? I think I can see the answer/reasoning, but I am not sure. It seems that this has to do with the fact that the initial concentrations are not necessarily equal and so you cannot just say that there are twice as much Y as X like with Ksp. But I don't know. Kaplan is **** for explanations and expects you to figure it out. Am I right in saying that's how they differ? and why is that so?

My brain exploded when I read how elaborate the second question was so I'll just pretend I never read it.

For the first question:

The Ksp is calculated by multiplying the concentration of ions of a presumably insoluble solid at equilibrium. Using the equilibrium reaction in your example:

XY2 <--> X + 2Y

The Ksp would be:

Ksp = [X][Y]^2 or [X][2X]^2

Both answers equal the same thing. In the second case you are just substituting X for Y since the concentration of Y equals 2X.

Now for Ion Product (Qsp):

When you use the ion product for a solubility question, you are usually trying to determine if a solution will have a precipitate given a certain concentration of ions that are not at equilibrium. The ion product tells you which direction the reaction will shift to maintain equilibrium (either forming more precipitate or more ions in solution).

Let's say the Ksp for XY2 is 1*10^-7, and you have a solution (not at equilibrium) where [X]= .15 and [Y] = .015. You want to know which direction the reaction will shift to maintain equilibrium (or whether it's already at equilibrium) so you use the Qsp to determine that.

So here Qsp equals:

Qsp = [X][Y]^2 = [.15][.015]^2 = whatever that equals.

Can't calculate it off the top of my head, but 1.5*10^-1 * 2.25*10^-4 is about 4*10^-5. In this case, Qsp > Ksp which tells you that there are more ions in solution than there should be, and the reaction will shift to the left making more precipitate. Conversely, if Qsp < Ksp, that means there are less ions in solution than there should be, and the reaction will shift to the right so that more ions dissociate into the solution. Finally, if Qsp = Ksp, you are at equilibrium.

Just as a summary, Qsp and Ksp are calculated the same way but are used to find different things. The Ksp tells you the amount of ions of an insoluble solid in solution at equilibrium. The Qsp is used to tell you which direction the reaction will shift based on the current concentration (or concentration at a given time) of the ions in solution.

 

[Kaustikos]

Okay, so let me see if I understand this;
the reason there's a multiplying factor is because we know that at equilibrium, the concentrations of the products are multiples of each other, BUT at non-equilibrium concentrations; this is not the case since we're dealing with initial concentrations.

So can I always assume that if we have an equilibrium reaction and are given the Ksp, we can assume that the concentrations of the products will be multiples of each other? Take this reaction

X2Y <----> 2X + Y

And we're given that Y = 4

Can I say that Ksp = [2 x 4]^2 [2], because this is at equilibrium and the concentration of X is half the concentration of Y?

 

[SketchLazy]

So can I always assume that if we have an equilibrium reaction and are given the Ksp, we can assume that the concentrations of the products will be multiples of each other? Take this reaction

X2Y <----> 2X + Y

And we're given that Y = 4

Can I say that Ksp = [2 x 4]^2 [2], because this is at equilibrium and the concentration of X is half the concentration of Y?

Yes. When one molecule of X2Y dissociates and the Xs from X2 aren't covalently bonded to each other or something like that, then you will always have a ratio of 2 X ions to 1 Y ion in solution at equilibrium. It will completely dissociate into its constituent ions if it dissociates, and you can't have a partial dissociation.

In your example,

Ksp = [2*4]^2[4]

 

[Kaustikos]

In your example,

Ksp = [2*4]^2[4]

stupid mistake

thanks

 

[phospho]

formation of a solution is exothermic, and we also happen to be increasing entropy while doing so.

changing h2o from liquid to solid is also exothermic but happens to be decreasing entropy.

sorry if it's a stupid question, but is it safe to assume that entropy has nothing to do with the fact that something is endothermic or exothermic? yes, it's probably exo if we increase entropy, but that's never definite...right?

thanks

 

[iA-MD2013]

formation of a solution is exothermic, and we also happen to be increasing entropy while doing so.

changing h2o from liquid to solid is also exothermic but happens to be decreasing entropy.

sorry if it's a stupid question, but is it safe to assume that entropy has nothing to do with the fact that something is endothermic or exothermic? yes, it's probably exo if we increase entropy, but that's never definite...right?

thanks

Yep...entropy has nothing to do with whether a process is exothermic or endothermic, but it does determine a reaction's spontaneity.

 

[RSAgator]

Yep...entropy has nothing to do with whether a process is exothermic or endothermic, but it does determine a reaction's spontaneity.

Just to make sure there's no confusion, entropy alone doesn't determine spontaneity but rather spontaneity is determined by the free energy equation.

dG = dH - TdS

For a given dH and dS reaction spontaneity is largely dependent on the temperature.

 

[phospho]

Yep...entropy has nothing to do with whether a process is exothermic or endothermic, but it does determine a reaction's spontaneity.

Just to make sure there's no confusion, entropy alone doesn't determine spontaneity but rather spontaneity is determined by the free energy equation.

dG = dH - TdS

For a given dH and dS reaction spontaneity is largely dependent on the temperature.

thank you both very much...

 

[iA-MD2013]

thank you both very much...

No problem

 

[phospho]

No problem

this is unrelated to this thread, but since it looks like you're awake right now, maybe you can help me

i'm having trouble drawing a mental picture of how it takes the exact amount of time for an elephant and a feather to hit the ground, assuming they fell from the same height

 

[iA-MD2013]

this is unrelated to this thread, but since it looks like you're awake right now, maybe you can help me

i'm having trouble drawing a mental picture of how it takes the exact amount of time for an elephant and a feather to hit the ground, assuming they fell from the same height

It's annoying, right? It's air resistance that makes the elephant falls much faster (greater acceleration) than the feather...but, in a vacuum, they would both have the same acceleration (g). Also, remember that the kinematics equations do not depend on mass, so the object's shape or size does not change anything in free fall, if there is no air resistance.

 

[phospho]

It's annoying, right? It's air resistance that makes the elephant falls much faster (greater acceleration) than the feather...but, in a vacuum, they would both have the same acceleration (g). Also, remember that the kinematics equations do not depend on mass, so the object's shape or size does not change anything in free fall, if there is no air resistance.

very nice, thank you...

 

[tncekm]

It's annoying, right? It's air resistance that makes the elephant falls much faster (greater acceleration) than the feather...but, in a vacuum, they would both have the same acceleration (g). Also, remember that the kinematics equations do not depend on mass, so the object's shape or size does not change anything in free fall, if there is no air resistance.

Just as a point of "pickyness" on my part, I think its worth stating that its the elephants inertia that makes the force of air resistance unable to change the elephants motion in the way that it does the feather, which has many orders of magnitude less inertia. I only bring it up b/c I hear the MCAT loves to talk of inertia qualitatively

 

[Kaustikos]

Just as a point of "pickyness" on my part, I think its worth stating that its the elephants inertia that makes the force of air resistance unable to change the elephants motion in the way that it does the feather, which has many orders of magnitude less inertia. I only bring it up b/c I hear the MCAT loves to talk of inertia qualitatively

Yeah, that's how I remembered it and how it basically helped explain the whole ball going up vs going down acceleration.

 

[phospho]

would someone be able to explain to me the reasoning behind buffer equations please?

for example, the question says that we want a buffer of pH=5, if HA has a pKa of 4.7.

I know we do 5=4.7 + log (A/HA)

if we do the math, we get (A/HA)=2

The answer in the book is "we add A- with one third equivalent of H3O".

I'm just confused. What did we do with the 2? How did they come up with the answer?

thanks!

 

[tncekm]

The discrepancy is b/c they want you to approximate. If pH ~ pKa you know that the [A]/[HA] is about 1.

But, anyway, to get a better idea of how to answer your question more entirely, can you post more information from the question / passage? Thx

 

[bluemonkey]

would someone be able to explain to me the reasoning behind buffer equations please?

for example, the question says that we want a buffer of pH=5, if HA has a pKa of 4.7.

I know we do 5=4.7 + log (A/HA)

if we do the math, we get (A/HA)=2

The answer in the book is "we add A- with one third equivalent of H3O".

I'm just confused. What did we do with the 2? How did they come up with the answer?

thanks!

If you add X of A- with 1/3X of H3O+, you will end up with 2/3X of A- and 1/3 of HA which gives you your A=2HA...

 

[tncekm]

If you add X of A- with 1/3X of H3O+, you will end up with 2/3X of A- and 1/3 of HA which gives you your A=2HA...

Or what he said

Just out of curiosity, which book is this? That required a calculator to get 10^0.3 = 2, and that's not very MCAT-like.

 

[phospho]

If you add X of A- with 1/3X of H3O+, you will end up with 2/3X of A- and 1/3 of HA which gives you your A=2HA...

thank you very much

Or what he said

Just out of curiosity, which book is this? That required a calculator to get 10^0.3 = 2, and that's not very MCAT-like.

berkeley review.

In the beginning of the book they emphasize that we must know that log 2=0.3 and log 3=0.48 and that those two can help us get the answer for any log question on the mcat. In this particular case, we wanted the inverse log of 0.3, which is 2. I think the book wanted to check that we remember that log 2=.3

thanks for your help!

 

[phospho]

in g=h-ts, i'm having trouble understanding why "t" is always positive. Why can't we have negative kelvin temperature?

thanks!

 

[tncekm]

0 K is defined as "absolute zero". This is where molecules have no internal kinetic energy and brownian motion of the molecules ceases. This is also the point where entropy is zero.

So, you can't really go below having zero energy, and you can't really have less than zero entropy. So 0 K, absolute zero, is as low as it goes!

Or something like that

For the MCAT be focused on the fact that at low values of T the spontaneity of a reaction will be dictated primarily by the enthalpy of the reaction (H) and at high values of T the spontaneity of a reaction will be dictated by the entropy changes. And, when H and S have opposite values its very easy figure out if the reaction will be spontaneous or not.

 

[phospho]

0 K is defined as "absolute zero". This is where molecules have no internal kinetic energy and brownian motion of the molecules ceases. This is also the point where entropy is zero.

So, you can't really go below having zero energy, and you can't really have less than zero entropy. So 0 K, absolute zero, is as low as it goes!

Or something like that

For the MCAT be focused on the fact that at low values of T the spontaneity of a reaction will be dictated primarily by the enthalpy of the reaction (H) and at high values of T the spontaneity of a reaction will be dictated by the entropy changes. And, when H and S have opposite values its very easy figure out if the reaction will be spontaneous or not.

thank you VERY much - makes perfect sense.

i have another question somewhat unrelated

I was reading some notes I took a few months ago, and I wrote:

"don't forget that it's the random kinetic energy of the molecules, not the uniform translational motion kinetic energy that increases the temp"

do you have any idea what i was talking about? lol

I just googled "uniform translational motion kinetic energy", and I couldn't find anything that would help me understand it.

sorry for asking so many questions. The real thing is in 3 days and I'm freaking out while trying to wrap up some loose ends.

thank you

 

[tncekm]

Yeah, I do

Okay, so, lets say a jug of water is sitting at 99C -- just before boiling. Very hot right? Well, how fast is it moving? Its translational speed is 0 m/s b/c its "sitting". The random "brownian" motion of the water molecules, however, is VERY fast. This random motion is what is what you're talking about.

This is why if you heat a reaction you can give it the energy it needs to overcome the activation energy. If you heat it, you increase the temperature, and you increase the random motion velocity, so the molecules have greater energy when they collide with each other. (Note: this is largely for endothermic reactions. always remember Le Chatliers take on "heat" as a product or reactant.)

Remember the equation KE = 3/2 RT this goes to show you that the internal kinetic energy is directly proportional to temperature.

Also, this random motion KE is also a major contributor to pressure. That's why if you increase temp in a fixed container containing a gas you increase pressure; the molecules are hitting the container with greater energy. This holds true for fluids, too. If you have a pipe that is 100% full of water, but its moving at 0 m/s, it has less pressure than if it were to move fast, right? This is because when you put energy into the uniform translational kinetic energy of the liquid system the energy has to come from somewhere, and if its not from an external source its coming out of the random internal kinetic energy.

 

[phospho]

Yeah, I do

Okay, so, lets say a jug of water is sitting at 99C -- just before boiling. Very hot right? Well, how fast is it moving? Its translational speed is 0 m/s b/c its "sitting". The random "brownian" motion of the water molecules, however, is VERY fast. This random motion is what is what you're talking about.

This is why if you heat a reaction you can give it the energy it needs to overcome the activation energy. If you heat it, you increase the temperature, and you increase the random motion velocity, so the molecules have greater energy when they collide with each other. (Note: this is largely for endothermic reactions. always remember Le Chatliers take on "heat" as a product or reactant.)

Remember the equation KE = 3/2 RT this goes to show you that the internal kinetic energy is directly proportional to temperature.

Also, this random motion KE is also a major contributor to pressure. That's why if you increase temp in a fixed container containing a gas you increase pressure; the molecules are hitting the container with greater energy. This holds true for fluids, too. If you have a pipe that is 100% full of water, but its moving at 0 m/s, it has less pressure than if it were to move fast, right? This is because when you put energy into the uniform translational kinetic energy of the liquid system the energy has to come from somewhere, and if its not from an external source its coming out of the random internal kinetic energy.

cool, i get it... now to find that random motion velocity, all we do is put KE=KE and make 1/2mv^2=3/2rt and solve for "v".

Out of curiosity (I might be be mixing up too many concepts together), can you tell me how that "v" is different from the "v" we get in {1/sq(molecular weight)} ? I know exactly when to use each equation, but it just occurred to me that the first "v" is called "velocity of a particle" and the second "v" is called "average speed of an atom or molecule". In a perfect world, would the two v's be equal, or am I talking about two completely different concepts?

thank you

 

[tncekm]

I'm not sure if that would work, so I can't answer you there.

The v in 1/sqt(MW) is related to the "translational" motion. I believe that comes from effusion, etc.

v1/v2 = sqt(m2/m1) <-- comes from this I believe, which comes from the average translational KE of particles in a gas will be the same, so you set 1/2mv^2=1/2mv^2 for two different types of particles and get their respective translational velocities.

 

[phospho]

I'm not sure if that would work, so I can't answer you there.

The v in 1/sqt(MW) is related to the "translational" motion. I believe that comes from effusion, etc.

v1/v2 = sqt(m2/m1) <-- comes from this I believe, which comes from the average translational KE of particles in a gas will be the same, so you set 1/2mv^2=1/2mv^2 for two different types of particles and get their respective translational velocities.

thank you so much for putting up with me!

 

[tncekm]

Not at a problem! Its a good way to refresh concepts, and when I type them out to explain them I have to think them through and make sure I understand them (or "think" I understand them) Ask away!

 

[Winterpegger]

The Kaplan review book says that bond breaking = endothermic, and bond formation = exothermic. From what (I think) I know, ATP --> ADP + Pi "releases" energy (the breaking of a "high-energy" phosphate bond; i also thought it was an exothermic reaction) Can someone please clear this up for me?

 

[tncekm]

Energy must be input to break the bond in ATP, but, because the new bonds made by ADP and Pi are more stable and are at a lower energy level. I.e. the Pi, aka PO4 3- is in a state of resonance that it couldn't achieve while still bonded to the adenosine molecule.

Look at the phosphates in ATP then look at phosphate by itself: http://guweb2.gonzaga.edu/faculty/cronk/biochem/images/phosphate-representations.gif

The bond between ADP and Pi is broken, and then they form new bonds.

 

[sleepy425]

The Kaplan review book says that bond breaking = endothermic, and bond formation = exothermic. From what (I think) I know, ATP --> ADP + Pi "releases" energy (the breaking of a "high-energy" phosphate bond; i also thought it was an exothermic reaction) Can someone please clear this up for me?

it's because you don't just break a bond in ATP hydrolysis. it's HYDROlysis. so you're right, if you were to simply break the phosphodiester bond, yeah, you'd have to just put in the energy. but it's hydrolysis, so you're forming a more stable bond between the phosphorous and the oxygen from the water (or the alcohol like in serine or threonine or tyrosine, or nitrogen of imidazole/histidine). anyway, so that new bond is more stable, partially due to resonance, partially because the phosphorous of a phosphate is electron poor, so the two other phosphates in ATP are electron withdrawing and make it less stable, so when you put an electron donating oxygen that is not attached to strong electron withdrawers, it's more stable.

soo, you're breaking a less stable bond and replacing it with a more stable bond, so breaking requires less energy than you're gonna gain from forming a more stable one. that's exactly why the Diels Alder rxn works too, same logic, you lose two C-C pi bonds (65 kcal/mol each) and replace them with two C-C sigma bonds (80 kcal/mol each) for a net gain of 30 kcal/mol. same reasoning.

 

[Farcus]

I have a question regarding half life.

Ok its a first order reaction with the equation .693/ak, with a being coeffient of reacant A and k being constant.

If the balanced equation is 4A+3B-> 2C with a rate constant of say .045

then the half life of C would be .693/(4)(.045) right?

I always thought that it was .693/k with "a" always being 1 but apparently not.

What volume of carbon dioxide would be
produced if 5 liters of propane, C3H8, were
burned in air according to the following
equation?
C3H8  5O2 &#10142; 3CO2  4H2O
A. 1 liter
B. 3 liters
C. 15 liters
D. 30 liters

how would you do this problem? i don't know where to begin except they're all gas and should be equaled but it gives me in Liters!!



Also if a drug is known to have second order reaction, then if a patient took the drug in huge quantitiy then the effect of the drug would go a lot more quicker than first order drugs right? since half life for second order is 1/ak[A]^0, where [A]^0 is the concentration of original substance.

 

[Kaustikos]

What volume of carbon dioxide would be
produced if 5 liters of propane, C3H8, were
burned in air according to the following
equation?
C3H8  5O2 &#10142; 3CO2  4H2O
A. 1 liter
B. 3 liters
C. 15 liters
D. 30 liters

how would you do this problem? i don't know where to begin except they're all gas and should be equaled but it gives me in Liters!!



Also if a drug is known to have second order reaction, then if a patient took the drug in huge quantitiy then the effect of the drug would go a lot more quicker than first order drugs right? since half life for second order is 1/ak[A]^0, where [A]^0 is the concentration of original substance.

Think of liters as a measurable quantity of gas for the sake of arguement. 5 liters of gas, we'll say, equals 5X of propane. Now, for every 1 part propane, we make 3 parts CO2. Thus, 1X = 3 CO2. So 5 x X = 5 x 3 CO2 = 15 CO2.

 

[Farcus]

clarrification on Collison Theory of Chemical Kninetics

so according to this theory the rate of any "SINGLE" step reaction such as CH3BR + OH is equal to the coefficient of that reaction so if the reaction was written as: 3 CH3BR + 2 OH -> Product, then the rate of reaction would be 5th overall with 3rd order methylbromide and 2nd order alcohol.

HOWEVER if its a multistep reaction in such as say

(CH3)3CBr(aq) + OH-(aq)

(CH3)3COH(aq) + Br-(aq)

​

then the rate of this reaction is equal to the SLOWEST STEP(

not shown here) and thus MEANING does NOT EQUAL TO THE coefficient of the BALANCED EQUATION if the reaction is multistage.



And this theory goes to support the law for rate law, in which [a]^x * ^y doesn't NECESSARILY mean the coeffieient of balanced equation for x and y, it only equals that if its single step reaction.


A question on Le Chatelier's Principle.
ok if a gas
2 SO2 is in dynamic equilibrium with O2, and an increase in volume which way would the reaction go? My reasoning for this question is, increasing volume = more moles of gas, and since there is 2 moles of SO2 for every 1 mole of O2, the reaction would go left to make the most of the volume. Would this be correct?

and if [SIZE=+1]H2(g) + 2ICl(g) <=> I2(g) + 2HCl(g), with increase in volume, there would be no overall change in the reaction right? going with similar reasoning.

also the only way to actually change equili constant Kc, is by increasing temp right? all other would only cause change in reverse or forward reaction right? such as concentration, and or pressure.
[/SIZE]​

 

[jae01]

I have a question regarding equilibrium.

What happens if you remove a reactant that is liquid? What happens in terms of equilibrium? Which way does it shift?

I thought it would have no effect since it's liquid. But Berkeley said that since you are removing a reactant, it shifts left.

Thanks!

 

[161927]

I have a question regarding equilibrium.

What happens if you remove a reactant that is liquid? What happens in terms of equilibrium? Which way does it shift?

I thought it would have no effect since it's liquid. But Berkeley said that since you are removing a reactant, it shifts left.

Thanks!

pure liquids do not affect le chatelier equilibrium (e.g., pure water)

aqueous reactants, however, do

 

[Farcus]

clarrification on Collison Theory of Chemical Kninetics

so according to this theory the rate of any "SINGLE" step reaction such as CH3BR + OH is equal to the coefficient of that reaction so if the reaction was written as: 3 CH3BR + 2 OH -> Product, then the rate of reaction would be 5th overall with 3rd order methylbromide and 2nd order alcohol.

HOWEVER if its a multistep reaction in such as say

(CH3)3CBr(aq) + OH-(aq)

(CH3)3COH(aq) + Br-(aq)

​

then the rate of this reaction is equal to the SLOWEST STEP(

not shown here) and thus MEANING does NOT EQUAL TO THE coefficient of the BALANCED EQUATION if the reaction is multistage.



And this theory goes to support the law for rate law, in which [a]^x * ^y doesn't NECESSARILY mean the coeffieient of balanced equation for x and y, it only equals that if its single step reaction.


A question on Le Chatelier's Principle.
ok if a gas
2 SO2 is in dynamic equilibrium with O2, and an increase in volume which way would the reaction go? My reasoning for this question is, increasing volume = more moles of gas, and since there is 2 moles of SO2 for every 1 mole of O2, the reaction would go left to make the most of the volume. Would this be correct?

and if [SIZE=+1]H2(g) + 2ICl(g) <=> I2(g) + 2HCl(g), with increase in volume, there would be no overall change in the reaction right? going with similar reasoning.

also the only way to actually change equili constant Kc, is by increasing temp right? all other would only cause change in reverse or forward reaction right? such as concentration, and or pressure.
[/SIZE]​

Can I get some answers for these questions thanks.

 

[budhia]

I'd use the RED CAT trick... just remember a red cat. Reduction is always at the cathode. That means that oxidation has to occur at the anode (you can think about AN OX for that).

A galvanic cell goes on its own... it doesn't need any outside work put in. An example is the battery that runs your cellphone or car. If it happens without work put in, that means it is spontaneous, so that means the sign of the Gibb's free energy is negative.

delta G = - n F E(cell)

This is a key formula for electrochemical cells. I'll refer to it often.

n is the number of electrons transferred (can never be negative), F is Faraday's constant (positive). The only way the sign of delta G can change is by the E(cell). We know a galvanic cell is spontaneous, therefore the delta G is negative. But we see on the right that there's already a negative sign. That means that the E(cell) must be positive to keep the delta G negative.

The inverse is true for an electrolytic cell. In an electrolytic cell we use work to drive an electrochemical reaction a way it doesn't want to go. That means that the reaction is nonspontaneous, since we have to add work to get it moving (plug it into a wall, add a battery, etc). Since it's nonspontaneous, delta G has to be positive. That means that the E(cell) must be negative in this case.

As far as reduction potentials, they're always given as the E(cell) for the half reaction. Let's look at two of them:

Hg(2+) + 2e- ---> Hg E=+0.85V
Zn(2+) + 2e- ---> Zn E=-0.76V

Ok, one's negative, and one's positive. What does that mean?

Well, lets refer back to the delta G formula. The mercury (Hg) reaction has a positive E, so that means the delta G for that reaction would be negative. That means that this is spontaneous, so it would be reduced without us having to do anything special.

As for the zinc, as it's written the E is negative, so that means the delta G is positive. That means that this reaction does NOT want to happen. So what we can say is:

For a galvanic cell, we know it's spontaneous. That means that we need both reactions to occur spontaneously.

That means that the mercury reaction can happen as written (so that reaction is good). Since mercury is reduced, that means it must be the cathode (remember RED CAT). Now, we need to balance the electrons, so we have to FLIP the other reaction. Zinc metal will be oxidized instead of reduced, so since we're running the reaction the other way, it is now spontaneous. (recall that if you flip a reaction, you flip the sign of the delta G).

Hopefully that kinda-sorta helps. It's a broad topic, so let me know where you are having specific trouble and I can help!

Nice job! with the above response. Can you shed how like on direction of the flow of current?

I know that electrons flow from anode to cathode! What about current? I would say current in the net flow of charge, so current flows in the direction that the electrons are flowing but apparently it is not so. Please let me know where I am missing the point.

Specific questions.
1. In concentration cell, which direction will the current flow
a) towards the lower concentration solution or
b) towards the higher concentration solution

2. acid- base concentration cell, which direction will the current flow
a) towards more acidic solution
b)towards more basic solution

please explain.

Thanks
b

 

[Chemist0157]

I've never been great with the electricity stuff, but I think current can be considered "positive flow" so it travels in the direction opposite of electrons.

I don't know how to answer #1, and my guess for #2 would be a more basic solution.

 

[ScoobyDooo]

I am sorry for reviving a 3 year old thread as I cannot find the justification for creating a new thread just for one question.


Can someone please help me understand why ammonium cation (NH4 +) cannot exist at a pH of 10. The Berk Review says it's because it has a pKa value less than 10. Was I supposed to know this information as background knowledge? The problem didn't provide a pka chart or any pertinent information leading to this conclusion.

Thanks in advance.