General Chemistry Question Thread

[QofQuimica]

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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.

 

[QofQuimica]

Sorry, I wasn't more specific, I meant between intramolecular bonds.
Polar, nonpolar, ionic, dipole-dipole, hydrogen bonds. I don't know the order of strongest to weakest. Do you?

I think you are having trouble distinguishing intermolecular interactions versus intramolecular bonds. There are two posts about them in the gen chem explanations thread that you might want to read if you haven't already. For intermolecular interactions, H bonds are the strongest, then dipole-dipole, then dispersion forces. For intramolecular bonds (ionic versus covalent), it depends. You can't make a blanket statement that ionic are always stronger than covalent or vice versa.

 

[pandora123]

Hi
Sorry - this a long question. I am getting terribly confused with pH and pKa problems.

I tend to think that strong acids dissociate fully and so pH = -log[H+].

For weaker acids, I calculate pKa = -log ([H+][A-]/[HA])
where HA is the initial acid concentration and A- is the conjugate base conc.
If x moles of HA are added to a 1 L solution then y moles of HA dissociate to produce y moles of H+ and y moles of A-.
So I calculate Ka = y^2/(x-y)

I recently came a across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Also, how do I know if the acid is weak or strong? If the question states a Ka or pKa value, can I assume the acid is weak?
Thanks a lot.

 

[gridiron]

Hi
Sorry - this a long question. I am getting terribly confused with pH and pKa problems.

I tend to think that strong acids dissociate fully and so pH = -log[H+].

For weaker acids, I calculate pKa = -log ([H+][A-]/[HA])
where HA is the initial acid concentration and A- is the conjugate base conc.
If x moles of HA are added to a 1 L solution then y moles of HA dissociate to produce y moles of H+ and y moles of A-.
So I calculate Ka = y^2/(x-y)

I recently came a across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Also, how do I know if the acid is weak or strong? If the question states a Ka or pKa value, can I assume the acid is weak?
Thanks a lot.

Hello. Your first statement about strong acids is correct--they do dissociate fully in water. Weak acids do not dissociate completely in water. For your question, set Ka= (moles H+)(moles conjugate base)/acid. What will help the most is the ICE chart (where I is the initial concentrations, c is change and e is equilibrium). Doing that, you see that you have the following: (0.0001 Molar acid, 0 moles H+ and conjugate base as initial concentrations. The changes is: 0.0001-x for acid, +x for H+ and conjugate base.) Now you have (x^2/0.0001-x)=Ka. Since this acid is weak, it dissociate's very moderatly in water so you can assume (this is the 5% approximation when the Ka is slightly greater than 10^-4) that 0.0001-x is so small that it is equal to 0.0001. Therefore, when you solve for x you do approximately get 8 * 10^-5.

For your other question. The strong acids are common (HCL, HBr, H2SO4, HI, HCLO4, HCLO3, and HNO3). These are the most common strong acids for which Ka>1. There are other acids that fit the definition of strong (thiocyanic acid, selenic acid, and permanganic acid) but are so uncommon that it is very unlikely to appear on the MCAT. I would suggest committing the above list to memory or that 3 are from the same group--the halogens. On the test, just because you are given a Ka value you cannot assume the acid is weak. You have to judge this by the magnitude of the Ka value. Strong acids will have a Ka value >>1 since they completely dissociate but weak acids will have a Ka value<<1 since they do not completely dissociate. Otherwise, I suggest to look at the identity of the acid--if it is from the list above, then it is a strong acid. I hope this helps and good luck

 

[QofQuimica]

Hi
Sorry - this a long question. I am getting terribly confused with pH and pKa problems.

I tend to think that strong acids dissociate fully and so pH = -log[H+].

For weaker acids, I calculate pKa = -log ([H+][A-]/[HA])
where HA is the initial acid concentration and A- is the conjugate base conc.
If x moles of HA are added to a 1 L solution then y moles of HA dissociate to produce y moles of H+ and y moles of A-.
So I calculate Ka = y^2/(x-y)

I recently came a across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Also, how do I know if the acid is weak or strong? If the question states a Ka or pKa value, can I assume the acid is weak?
Thanks a lot.

1) Yes. You won't have a calculator on the MCAT, so always make the approximation that y is very small compared to x and can be neglected in the denominator. That way, you will not have to ever use the quadratic equation. I got the same answer that you did.

2) You should memorize the list of strong acids for the test. Go to the gen chem explanations thread and read the post about strong acids there. Any acid not on the list should be assumed to be weak.

 

[shrutika]

Hi, I was wondering if someone can help me answer this question:

A 3:1 mixture of H2 and N2 was rapidly pumped into a Haber apparatus until the internal pressure was 60atm at 350 degrees celsius. A catalyst was then added and the system was allowed to reach equilibrium. The internal pressure was 45atm. What was the partial pressure of NH3 in the final mixture at 350 degrees celsius?

Answer - 15atm.

Thank you!

 

[naiominonna]

So I was wondering,

a question asks, which of the ions is expected to be most resitant to a reducing agent during reverse osmosis.

Choices:

Ni 2+
Cu 2+
Zn 2+ (This is the answer)
Ga 2+

In the passage it states tht reverse osmoiss, contains a substance that reduces the contaminants and that the ion easily gains electrons

No sure why the right answer is right

Thanks

 

[gridiron]

So I was wondering,

a question asks, which of the ions is expected to be most resitant to a reducing agent during reverse osmosis.

Choices:

Ni 2+
Cu 2+
Zn 2+ (This is the answer)
Ga 2+

In the passage it states tht reverse osmoiss, contains a substance that reduces the contaminants and that the ion easily gains electrons

No sure why the right answer is right

Thanks

Hey. I'm curious, do they provide a table of reduction potentials in the passage? If they do, this problem becomes a little bit easier. A reducing agent is a substance that gets oxidized during the course of the reaction. Normally, you would be give a table of reducing potentials in the passage. The more negative the reducing potential, the stronger the reducing agent the metal is. The best way I memorized the trend was: (and it may not be right but it helped me memorize it regardless) the more positive the reduction potential, the more favorable the species to picking up electrons. The more negative the reduction potential, the opposite reaction dominates. If you look at a table of reduction potentials, zinc has the most negative potential and thus is the more favorable reducing agent. I hope this helps!! Good luck.

 

[gridiron]

Hi, I was wondering if someone can help me answer this question:

A 3:1 mixture of H2 and N2 was rapidly pumped into a Haber apparatus until the internal pressure was 60atm at 350 degrees celsius. A catalyst was then added and the system was allowed to reach equilibrium. The internal pressure was 45atm. What was the partial pressure of NH3 in the final mixture at 350 degrees celsius?

Answer - 15atm.

Thank you!

Hey. The reaction between hydrogen gas and nitrogen gas produces ammonia. The balanced reaction looks like this:
3H2 + N2 --> 2NH3.
Since you are given molar quantities of two species, you know you are dealing with a limiting reactant problem. The stoichiometric ratio of H2 to N2 is 3:1 and the mixture is 3:1. That means nitrogen gas is the limiting reactant in the problem. One mole of nitrogen gas will produce 2 moles of ammonia. The reactant mixture contains a total of four moles. The partial pressure of H2 is (3/4)60 and N2 is (1/4)60. Since nitrogen is limiting, then this will give rise to a partial pressure of 15atm of ammonia gas in the product since all nitrogen gas is used up in the reaction. The remainder is hydrogen gas. I hope this helps. Good luck!

 

[pandora123]

1) Yes. You won't have a calculator on the MCAT, so always make the approximation that y is very small compared to x and can be neglected in the denominator. That way, you will not have to ever use the quadratic equation. I got the same answer that you did.

2) You should memorize the list of strong acids for the test. Go to the gen chem explanations thread and read the post about strong acids there. Any acid not on the list should be assumed to be weak.

Hi Everyone,
Thanks for your help. I have one follow up question -

In the above problem, the starting concentration of benzoic acid is 10^-4 moles/L. About 0.8x10^-4 moles/L of the acid dissociates and 0.2x10^-4 moles/L are left. Can this be considered a negligible concentration? Is there a particular cut-off below which the remaining acid concentration can be neglected on the MCAT?

 

[pandora123]

I recently came across the following situation: 0.0001 moles of benzoic acid (Ka = 6.6 x 10^-5) in 1L of water.
Solving this problem using the above method results in a quadratic equation. Should I assume that insignificant amount of acid will dissociate and so I can ignore the value y in the denominator? If I do this then y turns out to be approximately 8 x 10^-5. Is this correct?

Hi Everyone,
Thanks for your help. I have one follow up question -

In the above problem, the starting concentration of benzoic acid is 10^-4 moles/L. About 0.8x10^-4 moles/L of the acid dissociates and 0.2x10^-4 moles/L are left. Can this be considered a negligible concentration? Is there a particular cut-off below which the remaining acid concentration can be neglected on the MCAT?

 

[QofQuimica]

Hi Everyone,
Thanks for your help. I have one follow up question -

In the above problem, the starting concentration of benzoic acid is 10^-4 moles/L. About 0.8x10^-4 moles/L of the acid dissociates and 0.2x10^-4 moles/L are left. Can this be considered a negligible concentration? Is there a particular cut-off below which the remaining acid concentration can be neglected on the MCAT?

Like BME said, y should generally be 5% or less of x in order for the approximation to hold true. That's the cutoff for problems you do in gen chem class, anyway. But even if y is greather than 5% of x like in this case, always make the approximation anyway on the MCAT. Remember that the MCAT is a timed test. You don't have a calculator, and you don't have all day to solve these problems. So you need to be expedient and estimate here, not get your answer accurate to the thousandths decimal place. You really, really don't want to be trying to solve quadratic equations with no calculator.

 

[Andrew99]

Could someone please give me a breakdown of these?

What does a low Ka indicate?
A low Kb
A high Ka
A high Kb
A low pKa
A low pKb
A high pKa
A high pKb

I can't seem to get any of them straight except that a low pKa is similar to a low pH and that means a strong acid. Thanks in advance.

 

[QofQuimica]

Could someone please give me a breakdown of these?

What does a low Ka indicate?
A low Kb
A high Ka
A high Kb
A low pKa
A low pKb
A high pKa
A high pKb

I can't seem to get any of them straight except that a low pKa is similar to a low pH and that means a strong acid. Thanks in advance.

Ka is the equilibrium constant for an acid dissociation. In other words, if you have an acid dissociating:

HA <-> H+ + A-

Then Ka = {[H+][A-]}/[HA]

If you look at this equation, you will see that Ka is very large if the acid dissociates a lot (i.e., if it is a strong acid). In other words, for a strong acid, [HA] is small (small denominator) or [H+] and {A-] are large (large numerator). A low Ka means that the acid is weak: there is more HA than dissociated products.

Taking the pKa means you are taking the negative log. That means that if your Ka is low, when you take the negative log of that, you will get a large number, and vice versa. So a weak acid will have a larger pKa, and a strong acid will have a smaller one. (if you're not convinced of this, try taking some negative logs of several numbers using a calculator, and you will see that larger numbers have smaller negative logs and vice versa.)

Kb is similar, except that it is the equation for a base accepting a proton (if we used the Bronsted-Lowry definition). So here is the equation:

B + H2O <-> BH+ + OH-

Kb is set up analogously to Ka, and therefore a low Kb means you have a weak base that does not readily dissociate water. As before, if Kb is low, then pKb will be high, and vice versa.

 

[naiominonna]

question?

Isn't the more negative the reduction potential the higher liklihood it gets oxidized

Problem

two inert electrodes connected to a power source placed in aqueous solution of NACl, in resulting electrolytic solution , which forms at the anode?
E^0
1.Cl2 = 1.42
2.02 = -1.23
3.H2 =+.83
4.Na = +2.71

Answer says:

oxidation potential of water (–1.23 V) the opposite of its reduction potential is less negative than
the oxidation potential of chloride (–1.42 V). Thus, oxygen will form at the anode, not chlorine.

thanks

 

[gridiron]

question?

Isn't the more negative the reduction potential the higher liklihood it gets oxidized

Problem

two inert electrodes connected to a power source placed in aqueous solution of NACl, in resulting electrolytic solution , which forms at the anode?
E^0
1.Cl2 = 1.42
2.02 = -1.23
3.H2 =+.83
4.Na = +2.71

Answer says:

oxidation potential of water (–1.23 V) the opposite of its reduction potential is less negative than
the oxidation potential of chloride (–1.42 V). Thus, oxygen will form at the anode, not chlorine.

thanks

Usually, when you are given two species, you will be given the reduction potentials. The more negative the reduction potential, the better the species is a reducing agent--it is oxidized.

 

[pezzang]

Why is it that at standard state for an element the standard Gibbs energy is zero?

delta G = delta H + T(delta S)
At standard state, delta H for an element is, by convention, zero. So delta H part becomes zero. As for delta S, it won't be zero becuase it's zero at 0K(absolute temp) and in equilibrium. So how come delta G is zero? Please correct me..... 🙂

 

[pezzang]

.

 

[pezzang]

Is it possible to say the reaction to be exothermic when a solution of NaCl and another solution of AgNo3 are mixed to form a precipitate? My reasoning would be because that bonds formed from the precipitate reaction is stronger than the bonds broken so the overall heat of reaction should be exothermic. That is, less heat was put into breaking the bonds than released by the formation of stronger bonds, which is the precipiate (in solid state as compared to aqueous).
Am I on the right track? Thanks!!

 

[QofQuimica]

Why is it that at standard state for an element the standard Gibbs energy is zero?

delta G = delta H + T(delta S)
At standard state, delta H for an element is, by convention, zero. So delta H part becomes zero. As for delta S, it won't be zero becuase it's zero at 0K(absolute temp) and in equilibrium. So how come delta G is zero? Please correct me..... 🙂

The second term is also zero because it's DELTA S. The entropy isn't changing. 🙂

 

[QofQuimica]

Is it possible to say the reaction to be exothermic when a solution of NaCl and another solution of AgNo3 are mixed to form a precipitate? My reasoning would be because that bonds formed from the precipitate reaction is stronger than the bonds broken so the overall heat of reaction should be exothermic. That is, less heat was put into breaking the bonds than released by the formation of stronger bonds, which is the precipiate (in solid state as compared to aqueous).
Am I on the right track? Thanks!!

Assuming that your analysis is correct about the relative bond strengths, what you said makes sense to me.

 

[susan_williams]

I agree to what pezzang posted. where did yu learn bout it?

 

[pezzang]

The second term is also zero because it's DELTA S. The entropy isn't changing. 🙂

Is the delta S = 0 because there is no heat (delta H = 0) absorbed or released by the system? Thanks!

 

[pezzang]

I agree to what pezzang posted. where did yu learn bout it?

What do you mean?

 

[pezzang]

Assuming that your analysis is correct about the relative bond strengths, what you said makes sense to me.

Is it wrong to assume that the product (precipitate + non-precipiate) will have stronger bonds that the reactants as i assumed in the precipiate question? i was wondering because you said "assuming my analysis is correct"....

 

[QofQuimica]

Is it wrong to assume that the product (precipitate + non-precipiate) will have stronger bonds that the reactants as i assumed in the precipiate question? i was wondering because you said "assuming my analysis is correct"....

I don't know whether it does or not. But if you're right about the relative bond strengths of the products and reactants, then the rest of your argument sounds good to me.

I imagine that your scenario is the more common one, but there might be exceptions, so that's why I'm hesitating to say that this will ALWAYS be the case. For example, when you dissolve salts into water, that's usually exothermic. But there are a few that are endothermic. Analogously, I don't know whether every reaction where two salts form a precipitate is exothermic, including the example that you mentioned. But I think you're right that most of them are, if not all of them. 🙂

 

[pezzang]

Thanks QofQuimica. Your response really helped my understanding of heat of reaction. I have one more question that's bothering me.
Here's the question:

When salt X is dissolved in water to form a 1 molar unsaturaed solution, the temp of the solution is seen to drop.

The explanation says that this is endothermic reaction since heat was absorbed from the surrounding but I don't get how. I mean the solution is the system, right? Then, decreasing temp releases heat from the system. If the soln is not a system but surrounding giving up heat, what would be the system?

 

[gridiron]

Thanks QofQuimica. Your response really helped my understanding of heat of reaction. I have one more question that's bothering me.
Here's the question:

When salt X is dissolved in water to form a 1 molar unsaturaed solution, the temp of the solution is seen to drop.

The explanation says that this is endothermic reaction since heat was absorbed from the surrounding but I don't get how. I mean the solution is the system, right? Then, decreasing temp releases heat from the system. If the soln is not a system but surrounding giving up heat, what would be the system?

Temperature and heat are not the same quantities. Temperature is a measure of the average kinetic energy the system has. Heat is a measure of the energy that is transferred into or out of a system. The solution is saturated, so not all solute has dissolved into the solvent. When they say the temperature of the solution drops, I believe they are talking about the solvent. What happens is that as the solute is dissolved, heat is absorbed by the system. The system is the dissolved solute and the surroundings is the solvent. As heat is absorbed from the surroundings, the solvent, the temperature of the system increases and the temperature of the surroundings decreases. This is because the energy is lost from the surroundings. Heat has to be put into the system--hence a endothermic process. I hope this helps!

 

[QofQuimica]

Thanks QofQuimica. Your response really helped my understanding of heat of reaction. I have one more question that's bothering me.
Here's the question:

When salt X is dissolved in water to form a 1 molar unsaturaed solution, the temp of the solution is seen to drop.

The explanation says that this is endothermic reaction since heat was absorbed from the surrounding but I don't get how. I mean the solution is the system, right? Then, decreasing temp releases heat from the system. If the soln is not a system but surrounding giving up heat, what would be the system?

Yeah, the terminology is confusing here. I would also call the solution the system, and I would explain it this way: the system has decreased in temperature because it has not absorbed enough heat from the surroundings to keep the temperature constant. Therefore, it must be an endothermic reaction, because if the surroundings do not supply heat, an endothermic reaction leads to a drop in the temperature of the system. Consider this: if the surroundings were able to "replace" the heat that the reaction "needs," the temperature could have been kept constant. But that didn't happen here. So we can conclude that the surroundings didn't provide enough heat, to compensate for this endothermic reaction. Does that help?

 

[pezzang]

Temperature and heat are not the same quantities. Temperature is a measure of the average kinetic energy the system has. Heat is a measure of the energy that is transferred into or out of a system. The solution is saturated, so not all solute has dissolved into the solvent. When they say the temperature of the solution drops, I believe they are talking about the solvent. What happens is that as the solute is dissolved, heat is absorbed by the system. The system is the dissolved solute and the surroundings is the solvent. As heat is absorbed from the surroundings, the solvent, the temperature of the system increases and the temperature of the surroundings decreases. This is because the energy is lost from the surroundings. Heat has to be put into the system--hence a endothermic process. I hope this helps!

Thanks BioMedEngineer, you have helped me alot lately. I have one objection to make from your explanation. I initially thought the way you did but I realized that dissolving does not only involve heat being absorved into the syste (solute) because it actually involves breaking the bonds between solutes and then forming new bonds with solvents (solvents-solutes). Hence, the overall heat does not necessarily have to be absorbed by the system (solute). That's where my own understanding of distinguishing system (=solute) and surrounding (solvent) got stuck... Well, maybe I'm wrong..but something to think about... 🙂

 

[gridiron]

Thanks BioMedEngineer, you have helped me alot lately. I have one objection to make from your explanation. I initially thought the way you did but I realized that dissolving does not only involve heat being absorved into the syste (solute) because it actually involves breaking the bonds between solutes and then forming new bonds with solvents (solvents-solutes). Hence, the overall heat does not necessarily have to be absorbed by the system (solute). That's where my own understanding of distinguishing system (=solute) and surrounding (solvent) got stuck... Well, maybe I'm wrong..but something to think about... 🙂

I agree completely, but with the problem, that was the explanation I thought would cover the answer. Energy needs to be absorbed when breaking bonds and released when forming new bonds. That is where the quantity of change in enthalpy comes into consideration. 🙂

 

[pezzang]

Yeah, the terminology is confusing here. I would also call the solution the system, and I would explain it this way: the system has decreased in temperature because it has not absorbed enough heat from the surroundings to keep the temperature constant. Therefore, it must be an endothermic reaction, because if the surroundings do not supply heat, an endothermic reaction leads to a drop in the temperature of the system. Consider this: if the surroundings were able to "replace" the heat that the reaction "needs," the temperature could have been kept constant. But that didn't happen here. So we can conclude that the surroundings didn't provide enough heat, to compensate for this endothermic reaction. Does that help?

Great!!! Thanks again for your help.. 😀

 

[pezzang]

I have a problem here...

"delta H is not equal to q at constant volume; only at constant pressure."

I was trying to prove with formulas that the statement above is correct but I seem to keep bumping into a bit of problem...

@ constant volume:
delta U = q + P(delta V) and here P(delta V) = 0 since no volume change

so, delta U = q.

Now delta H = delta U + P(Delta V) and again P(delta V) = 0 for the same reason as above.

Hence, we can say that delta H = delta U = q. So how come delta H is not equal to q at const volume? Sorry for another full of algebra question... 😀

 

[gridiron]

I have a problem here...

"delta H is not equal to q at constant volume; only at constant pressure."

I was trying to prove with formulas that the statement above is correct but I seem to keep bumping into a bit of problem...

@ constant volume:
delta U = q + P(delta V) and here P(delta V) = 0 since no volume change

so, delta U = q.

Now delta H = delta U + P(Delta V) and again P(delta V) = 0 for the same reason as above.

Hence, we can say that delta H = delta U = q. So how come delta H is not equal to q at const volume? Sorry for another full of algebra question... 😀

I'll try to take a crack at this....changes in enthalpy occur at constant pressure--open system. A good way to understand what enthalpy means is through an analogy. Say we have a system--nothing is in the system but we give in some internal energy. Since we have given the system internal energy room has to be made to give space to the system. So, if the pressure is held constant, in this case atmospheric pressure, than work must me done by system against the atmosphere. Conversly, if energy is removed by the system, the work is done on the system by atmosphere. Therefore, enthalpy changes at constant pressure and I think it is correct to say the formula is: delta H = delta U + PdeltaV. I believe work in the form of compression and expansion can be ignored. Since Q= delta U + PdeltaV, than delta H is also equivalent to delta Q. I hope this helps and good luck

 

[pezzang]

I'll try to take a crack at this....changes in enthalpy occur at constant pressure--open system. A good way to understand what enthalpy means is through an analogy. Say we have a system--nothing is in the system but we give in some internal energy. Since we have given the system internal energy room has to be made to give space to the system. So, if the pressure is held constant, in this case atmospheric pressure, than work must me done by system against the atmosphere. Conversly, if energy is removed by the system, the work is done on the system by atmosphere. Therefore, enthalpy changes at constant pressure and I think it is correct to say the formula is: delta H = delta U + PdeltaV. I believe work in the form of compression and expansion can be ignored. Since Q= delta U + PdeltaV, than delta H is also equivalent to delta Q. I hope this helps and good luck

Thanks again. I have a question though. How come we can assume that work in the form of compression and expansion (=PV work) can be ignored? And I think EK said that q is a transfer of energy so delta q can't exist. Also at constant pressure, q is not equal to delta U.
I was trying to prove why delta H is not equal to q at constant volume because i thought if there is no change, there can be no PV work and hence delta U = q = delta H (= delta U + P*delta V(=0))....... Any help would be appreciated!! 🙂

 

[gridiron]

Thanks again. I have a question though. How come we can assume that work in the form of compression and expansion (=PV work) can be ignored? And I think EK said that q is a transfer of energy so delta q can't exist. Also at constant pressure, q is not equal to delta U.
I was trying to prove why delta H is not equal to q at constant volume because i thought if there is no change, there can be no PV work and hence delta U = q = delta H (= delta U + P*delta V(=0))....... Any help would be appreciated!! 🙂

Regarding the first question, why compression and expansion, I'm not too sure of it myself because I read it one of my chemical engineering books. And yes you are right, it was a typo on my part, it isn't delta q. Regarding the other question....
Consider the two types of calorimetry: constant volume and constant pressure. At constant volume, no work is peformed so any heat measured equals changes in the system's internal energy. At constant pressure, q is equal to the change in internal energy minus the work performed. Any heat measured represents the change in enthalpy. Consider this point: we have a system open to the atmosphere. We then give sufficient energy, internal energy, to the system. The problem is that we need to make room for the system since it has gained internal energy. Work needs to be done againt the atmosphere---at constant pressure. This work and the gain in internal energy represents the change in enthalpy. At constant volume, no work is peformed on the system or by the system, so any changes are due to pressure, internal energy or energy moving in the form of heat. Enthalpy can only be taken into consideration at constant pressure. That is why the heat of fusion and vaporization are given standard values at specific pressures. When you heat something at constant pressure, you allow it to expand. I hope this helps!!!

 

[Gabby]

Can someone please explain the Born-Haber cycle to me? I feel like I just memorized it for my Chem I test on it and even though I'm done with it as far as class is concerned, I don't feel like I truly grasped it. I understand Hess's Law, but my problem is writing up the equations. For example, KCl would be:

K+ (s) ----> K(g) + e-

One of my books has it exactly the opposite, but my class textbook has it the way I wrote it above. Which is right or does it not matter as long as you write the Cl in a way where the electrons cancel out? I just don't get the equations. If anyone can help me, I'd greatly appreciate it.

 

[pezzang]

why is "the strongest bonds present in a glass of water" and "the strongest bonds MADE by water (molecules) in a glass of water" different? Why is that even though water molecules CONSIST of covalent H-O bonds, they do not MAKE covalent bonds (with other molecules) but they make hydrogen bonds?

 

[Dr Durden]

An oxygen atom has six electrons initially but needs two more electrons for a full valence octet. To get the final two, it shares them with two hydrogen atoms-which simply have one electron apiece-in two very strong covalent bonds. Neither species ionizes though as the difference in electronegativity is not great enough. However, a difference in electronegativity does still exist. As the lone oxygen atom much more electronegative, it attracts electrons to its nucleus more so than the hydrogen atoms. This leads to the oxygen atom having more of a transient negative charge and the hydrogen atom with more of a transient positive charge. The oxygen atom has its full valence octet and can no longer form covalent bonds, but the negative oxygen atom is still attracted to positive hydrogen atoms on nearby water molecules. The INTRAmolecular covalent bond is strong taking about 460 kJ/mol to disrupt, but these transient charge differences lthus ead to weaker INTERmolecular hydrogen bonds of about 20 kJ/mol.

 

[DCDAWG]

Quick Question:

Metals in their elemental state cannot be reduced, only oxidized- correct?

 

[gridiron]

Quick Question:

Metals in their elemental state cannot be reduced, only oxidized- correct?

Not necessarily in my opinion. What you should look at is the reduction potentials given to you or any chart/graph that might illicit what is being oxidized in solution or reduced. Good luck!

 

[DCDAWG]

Not necessarily in my opinion. What you should look at is the reduction potentials given to you or any chart/graph that might illicit what is being oxidized in solution or reduced. Good luck!

Well thats certainly the obvious and logical explanation. Unfortunately, you would have made the same mistake as I on kaplan physics section test #4.....

A passage stated that students were given their choice of silver, copper, tin, nickel, and iron to react with HCl to compare their reactivities. The reduction potentials of each metal were given at .8, .34, -.14, -.25, and -.44; respectively.

Obviously tin, copper, and iron turn out to be the most reactive, with their negative reduction potentials, they will be the most easily oxidized to their metal salt CuCl2, etc, and will reduce H+ to H2 gas.

What confused me, is in the explanation of the passage and some of the questions, kaplan states that metals in their ground or elemental state can only be oxidized, not reduced. They will much more readily lose electrons (in this case acting a reducing agent for H+) than gain electrons.

This is confusing because I would expect the metals listed with positive reduction potentials to have the ability to be reduced.

Perhaps QofQuimica could elaborate....

Thanks.

 

[xanthomondo]

Well thats certainly the obvious and logical explanation. Unfortunately, you would have made the same mistake as I on kaplan physics section test #4.....

A passage stated that students were given their choice of silver, copper, tin, nickel, and iron to react with HCl to compare their reactivities. The reduction potentials of each metal were given at .8, .34, -.14, -.25, and -.44; respectively.

Obviously tin, copper, and iron turn out to be the most reactive, with their negative reduction potentials, they will be the most easily oxidized to their metal salt CuCl2, etc, and will reduce H+ to H2 gas.

What confused me, is in the explanation of the passage and some of the questions, kaplan states that metals in their ground or elemental state can only be oxidized, not reduced. They will much more readily lose electrons (in this case acting a reducing agent for H+) than gain electrons.

This is confusing because I would expect the metals listed with positive reduction potentials to have the ability to be reduced.

Perhaps QofQuimica could elaborate....

Thanks.

Have you ever seen a metal anion?

Remember that the standard reduction potentials are all numbers that are relative to hydrogen's standard reduction potential...just because the value is positive doesn't mean that it can be an oxidizing agent

 

[mcat_study]

In question # 59 of the AAMC7,
why is the answer
a saturated solution with 1.3g of undissolved salt instead of supersaturated solution (answer D). I looked up a definition of a supersaturated solution on wiki and in my mind, it describes exactly what the questions stem is asking. maybe the key is that it was left over for several days and the solute adsorbed to a crack or got nucleated?
please help
thank you 🙂

 

[DCDAWG]

Have you ever seen a metal anion?

Remember that the standard reduction potentials are all numbers that are relative to hydrogen's standard reduction potential...just because the value is positive doesn't mean that it can be an oxidizing agent

Right.

 

[nope80]

Hi everyone,

This is a great thread!

I have a few questions about some material I encountered in a recent TPR workbook.

1. What is the relationship between energy, wavelength and frequency? This was in the chemistry section.

2. Sp3. I thought that sp3 refers to an atom with 4 singly bonded atoms OR nonbonding e-. So like CH4 is sp3 and same with R-O-R. But I guess the latter is incorrect? And that nonbonding electrons do not count??

3. Electron Affinity. Does this refer to the amount of energy needed to add an electron or the amount released after an electron is added? So in other words if EA is negative does that mean the atom wants the electron or not?

Thanks a lot!🙂

 

[MSc44]

how can you break a rnx down into steps?????

ex NO2(g) + CO(g)---> NO(g) + CO2(g)

i know what the slow and fast steps are, but HOW do you derive them???

 

[QofQuimica]

In question # 59 of the AAMC7,
why is the answer
a saturated solution with 1.3g of undissolved salt instead of supersaturated solution (answer D). I looked up a definition of a supersaturated solution on wiki and in my mind, it describes exactly what the questions stem is asking. maybe the key is that it was left over for several days and the solute adsorbed to a crack or got nucleated?
please help
thank you 🙂

mcat_study:

1) Please don't start new threads in the MCAT subforum. Just post your question in the appropriate already existing thread.

2) I don't have the AAMC tests, so I can't answer your question. We are not supposed to post AAMC questions on SDN, but if you can ask about the general concept without posting the specific question, we'll do our best to help. 🙂

 

[xanthomondo]

how can you break a rnx down into steps?????

ex NO2(g) + CO(g)---> NO(g) + CO2(g)

i know what the slow and fast steps are, but HOW do you derive them???

Its not possible to derive reaction mechanisms...you can only guess and run into the lab to experiment and see if you're right

on the MCAT youll always be given the mechanism

 

[MSc44]

Its not possible to derive reaction mechanisms...you can only guess and run into the lab to experiment and see if you're right

on the MCAT youll always be given the mechanism

thanks i see what your saying



can i not post questions here??????

 

[xanthomondo]

Hi everyone,

This is a great thread!

I have a few questions about some material I encountered in a recent TPR workbook.

1. What is the relationship between energy, wavelength and frequency? This was in the chemistry section.

2. Sp3. I thought that sp3 refers to an atom with 4 singly bonded atoms OR nonbonding e-. So like CH4 is sp3 and same with R-O-R. But I guess the latter is incorrect? And that nonbonding electrons do not count??

3. Electron Affinity. Does this refer to the amount of energy needed to add an electron or the amount released after an electron is added? So in other words if EA is negative does that mean the atom wants the electron or not?

Thanks a lot!🙂

1. E = hv= hc/(lambda) (v is pronounced "new"..its the frequency)

2. Ethers are sp3...it has no double bonds...non bonding electron pairs do count...the electron pair geometry is still tetrahedral

3. Electron affinity shows the amount of energy lost or gained when a neutral molecule gains an electron (its hard to measure)...it being negative only means that it releases energy as it captures the electron...so negative delta Hs would mean its favorable... it doesnt "want" anything lol

be careful with Ea...thats usually used for activation energy of a reaction...the EA you're describing is really delta H