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

*********

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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[HristosKaran]

Hey everyone,

Rh(Ph3P)3Cl + H2 <--> Rh(P3)3H2Cl

Can someone just explain to me how to find the formal oxidation state of this reaction for rhodium? The reaction is in equilibrium... The answer said the rhodium species experiences an increase in formal oxidation state, but wouldn't the excess H2's lead to a reduction? Thanks

 

[BrokenGlass]

Hey everyone,

Rh(Ph3P)3Cl + H2 <--> Rh(P3)3H2Cl

Can someone just explain to me how to find the formal oxidation state of this reaction for rhodium? The reaction is in equilibrium... The answer said the rhodium species experiences an increase in formal oxidation state, but wouldn't the excess H2's lead to a reduction? Thanks

Electronegativities below are from from http://en.wikipedia.org/wiki/Pauling_Electronegativity_Scale

Rh is 2.28
C is 2.55
P is 2.19
Cl is 3.16
H is 2.20

In the reactant, Rh is bonded to 3 P atoms and one Cl atom. P is less electronegative than Rh while Cl is more electronegative
than Rh, so oxidation state of Rh in the reactant is -3 + 1 = -2.

In the product, Rh is bonded to 3 P atoms, 2 H atoms, and 1 Cl atom. H is less electronegative than Rh. So oxidation state of Rh
in the product is -3 + (-2) + 1 = -4.

So looks like oxidation state of Rh decreases.

We didn't even have to do the math. Since Rh gains 2 bonds to H atoms (while all other bonds between Rh and the atoms to which
it's attached remain the same), Rh must have been reduced... and reduction is always accompanies by a decrease in oxidation state (because electrons are gained and electrons are negative).

Note that Rhodium is a transition metal and can break the octet rule (which it does in the product).

Btw, where is this problem from? I am curious as to why I am not getting the same answer as the answer key you provided....

 

[HippocratesX]

I don't know if someone can help me with an actual MCAT practice question on AAMC #3 exam?

I having difficulty understanding question #35 on the exam. It's a question relavent to molality. Although I've read the answer, I'm still having trouble comprehending how they calculated that oxalic acid solution provided the greatest number of moles of solute as compared to the other solutions.

I was wondering if anyone can give a step by step thought process explanation...

I think the key for this kind of question, is not necessarily knowing the details of pKa in relation to freezing point depression....but instead, to know the equation for the colligative properties; in this case of freezing point depression.. Delta(Tf) = k x molality. And knowing that molality is (mols/1kg)...u can see that its a concentration of sorts. The 4 answer choices are in terms of mass percent or molar concentration. Mass percent is not a concentration...so those 2 answer choices can be crossed out.

Next, by looking at the equation for Tf...u can see that as the molality (or in our present case molarity or concentration increases)...so does the freezing point depression. The question asks which one has a lower freezing point. And freezing point is a colligative property, which means it only depends on how many solutes u have in solution...doesnt matter what kind they are. So from this, we can see the one with greatest concentration has the biggest dip or depression in freezing point....and the only answer choice that says soemthin has a greater concentration is answer choice C.

So the more solutes u have in solvent...the greater molality and molarity u will have. it just depends the language u are using to measure them (moles/Liter or moles/kilogram). And the greater

 

[rox1co]

so i'm doing a lot of chem prob. this summer bc i will be tutoring upcoming Fall, i'm having trouble with this one problem

A mixture containing KClO3, K2CO3, KHCO3, and KCl was heated, producing CO2, O2, and H2O gases according to the following reactions:
2KClO3(s) &#8594; 2KCl(s) + 3O2(g)
2KHCO3(s) &#8594; K2O(s) + H2O(g) + 2CO2(g)
K2CO3(s) &#8594; K2O(s) + CO2(g)
The KCl does not react under the conditions of the reaction. If 100.0 g of the mixture produces 1.80 g of H2O, 13.20 g of CO2, and 4.00 g of O2, what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

i've successfully gotten the amount of KClO3 & KHCO3; however, K2CO3 is giving me trouble (i'm thinking that maybe K2O is the limiting reagent, but how do i get that?)

*point the arrow the other way to make it look like a regular Limiting Reactant problem.

answers: KClO3 = 10.2g, KHCO3 = 20.0g, K2CO3 = 13.8g (eh? idk why i'm not getting it), KCl does not react so it's just 100g - (KClO + KHCO3 + K2CO3) = 56.0g

thanks for any help

 

[pezzang]

What is the difference between the heat of solution and heat of formation? Heat of formation is the chang in enthalpy when a compound is formed from its elementary forms. The heat of solution is the sum of enthalpies when a solution is formed. I can't see what's different... Any help would be appreciated! Thanks!

 

[rcd]

What is the difference between the heat of solution and heat of formation? Heat of formation is the chang in enthalpy when a compound is formed from its elementary forms. The heat of solution is the sum of enthalpies when a solution is formed. I can't see what's different... Any help would be appreciated! Thanks!

When you dissolve NaCl, you get Na+(aq) and Cl-(aq). When you calculate NaCl's enthalpy of formation, you use Na(s) and .5Cl2(g). Both the phase and the charges are different, which means enthalpies are different.

edit: it's .5 Cl2(g), not Cl(s) like I wrote intitially.

 

[pezzang]

Thank you, rcd. One more question about partial pressure and vapor pressure of a liquid. Evaporation occurs when the liquid's pp is less than its vp but the Patm is greter than the vp. Here, I realize vp is the vp created by the random translational KE of the liquid, but can't seem to distinguish the differnece between the pp and vp of a liquid. Any help would be appreciated!

 

[rcd]

Thank you, rcd. One more question about partial pressure and vapor pressure of a liquid. Evaporation occurs when the liquid's pp is less than its vp but the Patm is greter than the vp. Here, I realize vp is the vp created by the random translational KE of the liquid, but can't seem to distinguish the differnece between the pp and vp of a liquid. Any help would be appreciated!

Pp is the current gas pressure due to the substance. Vp is the equilibrium Pp of the substance.

So, if you put liquid water into an rigid container with N2 gas at 5atm and close it, you start with 0 Pp H2O. Vp will be some constant dependent on temperature and heat of vaporization. If you wait long enough, enough H2O will evaporate and Pp will equal Vp. If you had no N2 gas to begin with, water would boil until Pp equalled Vp.

 

[xlr8]

Hi, I need some clarification/help on this question:

Ammonia burns in air to form nitrogen dioxide and water.

4NH3 (g) + 7O2 (g) --> 4NO2 (g) + 6H2O (l)


If 8 moles of NH3 are reacted with 14 moles of O2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO2 in the container when the reaction runs to completion? (Assume constant temperature)

Since there is no limiting reagent (4:7 holds for both reactants that are put in), I can assume 8 moles of NO2 and 12 moles of H2O are created, or 4 and 6 respectively to correspond to the equation. Now I know that the partial pressure is the mol fraction times the total pressure, but my question is do I include the H2O in the mol fraction since it is a liquid and not a gas? Thus, is the answer (4/10)*(11 atm) ?

 

[philios]

Hi Im going through the examkrackers Chemistry book, and came upon a study question I dont understand...(its #22 if that helps)
22)Which of the following best explains why sulfur can make more bonds than Oxygen?
a)Sulfur is more electronegative than oxygen
b)oxygen is more electronegative than sulfur
c)sulfur has 3d orbitals not available to oxygen
d)sulfur has fewer valence electrons

They say the answer is C, which confuses me..I thought sulfur was
[Ne]3s(^2)3p(^4) , which does not include 3d at all! Is it because sulfur has the potential to use the 3d orbital? Thanks!

 

[philios]

Hi, I need some clarification/help on this question:

Ammonia burns in air to form nitrogen dioxide and water.

4NH3 (g) + 7O2 (g) --> 4NO2 (g) + 6H2O (l)


If 8 moles of NH3 are reacted with 14 moles of O2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO2 in the container when the reaction runs to completion? (Assume constant temperature)

Since there is no limiting reagent (4:7 holds for both reactants that are put in), I can assume 8 moles of NO2 and 12 moles of H2O are created, or 4 and 6 respectively to correspond to the equation. Now I know that the partial pressure is the mol fraction times the total pressure, but my question is do I include the H2O in the mol fraction since it is a liquid and not a gas? Thus, is the answer (4/10)*(11 atm) ?

im looking at the exact same problem in my examkrackers chem book (problem #27). The way they explain it is that since you have an initial pressure of 11, then each gas is contributing a partial pressure equal to their stoichometric coefficient (# of moles). Since it runs to completion and there are 4 moles of NO2 in the product, the partial pressure will be 4 moles. Water is NOT included since it is a liquid, and if there were a solid product that would not be included either (only those products that have a (g) after them).

 

[xlr8]

im looking at the exact same problem in my examkrackers chem book (problem #27). The way they explain it is that since you have an initial pressure of 11, then each gas is contributing a partial pressure equal to their stoichometric coefficient (# of moles). Since it runs to completion and there are 4 moles of NO2 in the product, the partial pressure will be 4 moles. Water is NOT included since it is a liquid, and if there were a solid product that would not be included either (only those products that have a (g) after them).

But since it runs to completion won't the only gas be NO2? To have those 4 moles of NO2 wouldnt you have used up your 4 and 7 moles of ammonia and oxygen?

And let's say I did go by their reasoning (which doesn't make sense to me because there cant be any reactants left while there is 4 moles of NO2), the mole fraction of NO2 would be (4/(4+7+4)), times 11, gives ~3 atm?

 

[philios]

11 moles is the INITIAL pressure within the container prior to the reaction. Since this initial pressure is 11 we can deduce that every mole of gas (7 + 4) contributes 1 to the pressure, therefore after the reaction since there is ONLY 4 moles of no2 within the system (like you mentioned), its partial pressure would be the whole pressure which would be 4. Hope this helps

 

[xlr8]

I see how initially, each mol for each gas contributes 1 atm, but I don't see how it can still apply after the react comes to completion.

For instance, say the initial pressure was 10 atm, what would the answer be then?

 

[rcd]

I see how initially, each mol for each gas contributes 1 atm, but I don't see how it can still apply after the react comes to completion.

For instance, say the initial pressure was 10 atm, what would the answer be then?

You'd multiply everything by 10/11.

 

[xlr8]

So generally, say you have a reaction:

aA + bB --> cC + dD , all are gases

The initial pressure is x atm.

to find the partial pressure of C:

(c / (c+d) ) * (x / (a+b) )

So basically you idealize it so that the initial pressure corresponds to the stoichiometric coefficients (sum of a and b) and then adjust it according to your initial pressure?

And if I was given the final pressure of the vessel I would do the same thing but reverse?

The trouble I was having was connecting the reactant partial pressure to the product partial pressure, but I think I have an understanding now:

You use the coefficients because there is no relation between initial pressure and final pressure of the container, so your stoichiometric coefficients are the relationships used to go from initial pressure conditions to final pressure conditions, correct?

 

[xlr8]

I have another question, this time regarding PV=nRT/kinetic molecular theory.

So, from what I've read I understand that you can only apply PV=nRT proportional relationships when you have 2 variables (hence, P,V, or T must be constant). Is this correct?

If all 3 are variable, then you must determine changes by looking at the internal energy.

Volume Changes
If the volume is increased, the gas does work on the surroundings via the kinetic energy, thus losing KE and T (since KE is related to T by 3/2kT). Pressure decreases because it is the KE per volume.

If the volume is decreased, the force does work on the gas, increasing the internal energy, thus the KE and the T? The pressure increases because the KE per volume is increased, from both the KE and volume.

Pressure Changes
Now here, I am hesitant to draw the same conclusions as above. If the pressure is increased/decreased, does that mean that the KE is altered or the volume, or would it have to be given that one remains constant?

 

[ocwaveoc]

Hi,
I've noticed that my Chem book states that CaCl2 is soluble in water. According to the SDN study thread, halogens with Ca are insoluble.
Am I mistaken?
Thanks.

 

[corbis11]

Hi,
I've noticed that my Chem book states that CaCl2 is soluble in water. According to the SDN study thread, halogens with Ca are insoluble.
Am I mistaken?
Thanks.

CaCl2 is soluble. I couldn't find anywhere on the study thread where it contradicts this.

 

[ocwaveoc]

CaCl2 is soluble. I couldn't find anywhere on the study thread where it contradicts this.

Thanks for the reply and clarification.
Under "General chemistry topic" thread and under "Mnemonic for salt solubility rule" it says:
Here is SilvrGrey330's extremely clever mnemonic to remember salt solubility rules.

C A S H n Gia

Read it as "Cashin' Gia"...how to remember that? well the story is...im a pimp...and gia is my hoe, and i need to get my cash from her. hence...Cashing from gia.

C is clorates, A is acetates, S is sulfates, H is halogens, n is Nitrates, and Gia is Group I A metals. ---> THESE ARE ALL SOLUBLE, XCEPT

for S: Ca, Ba, Sr.....just remember the tv network CBS
for H: Ca, Ba, Sr + Happy...whats happy? Hb Ag Pb ...mercury, silvr and lead...add a py to the end and all the first letters spell HAPPY

and if its not part of CASHnGIA...its insoluble.

**********

Please note that salt solubility is a complex equilibrium, and it isn't an either-or phenomenon like a simplified set of rules makes it out to be. Books may differ on how detailed they get about the rules for predicting it, and also on where they draw the line between calling a salt "soluble" versus calling it "insoluble." Keep in mind that the solubility rules are guides to help you predict the relative solubilities of different salts, and take them all with a grain of (ahem!) salt.

Doe that say all halogens are soluble except "Ca, Ba, Sr and Happy"?

Thanks...I'm probably misreading something.

 

[rox1co]

i think mnemonics suck, study hard!

 

[IDBasco]

For a given exothermic reaction, how are delta H, Ea (forward), and Ea (reverse) related?

I say that its delta H = Ea Forward - Ea Reverse, but my thing here says its delta H + Ea Forward = Ea reverse.

Wouldn't delta H have to be positive in order for their answer to be correct? Which would make the reaction not exothermic...

I visualize this reaction in the way we learn reaction coordinates in organic. Product side (on right) of the chart is lower H than the reactants (on left) with a hump in the middle. Am I totally out in left field here?

 

[rcd]

For a given exothermic reaction, how are delta H, Ea (forward), and Ea (reverse) related?

I say that its delta H = Ea Forward - Ea Reverse, but my thing here says its delta H + Ea Forward = Ea reverse.

Wouldn't delta H have to be positive in order for their answer to be correct? Which would make the reaction not exothermic...

I visualize this reaction in the way we learn reaction coordinates in organic. Product side (on right) of the chart is lower H than the reactants (on left) with a hump in the middle. Am I totally out in left field here?

You're right.

 

[IDBasco]

You're right.

Yeah, I should have just waited for Kaplan to get back to me. They agree with you, and me. Thanks though.

 

[mdfresh]

So here is the problem. All of you probably know the answer to this and I feel kinda dumb asking it becuase it seems like it would be easy but not so much for me. I was working in my mcat review book when the following conversion came up. Convert 12.6h to ms. No big deal except they set up the problem like so.. (12.6)(3,600g/1h)(1ms/10-3g)= 4.54*E7ms... What in the world does the g stand for. I have looked on the internet but I cannot find an explanation. Thanks for the help guys.

 

[corbis11]

Has to be a typo

They have it setup correctly, but the "g" should be s for seconds

 

[mdfresh]

Thats what I thought so I am glad someone else thinks so to. I sat there forever trying to figure out what in the world that could stand for. Anyways thanks for the help.

 

[bluediamond]

the answer to this is probably way easier than i am making it out to be but here goes...

on a practice exam, i was trying to draw NO3- and ended up making it O=N-O-O because this way there is only one oxygen with a negative charge (on the right). but apparently it should be O-N(=O)-O... so that two oxygens are negative and the nitrogen is positive. it seems really counterintuitive to me that there would be 3 charged atoms instead of 1. can someone explain why this is? also, is there a way to figure out which atoms are connected to each other and how to draw structures (other than counting free electrons because this didn't help me)? thank u!

 

[corbis11]

that negative charge is localized so it is unstable. The correct way allows for the charge to be delocalized evident by the three resonace structures one can draw using O-N(=O)-O

 

[bluediamond]

thanks corbis11! i didn't even think about resonance

 

[bluediamond]

another question...

rxn 1: 2I (-) + S2O8 (2-) --> I2 + 2SO4 (2-)

there is a table with rxn rates at various concentrations of the reactants and at different temperatures. when CuSO4 is added, the rxn time is cut in half, and the question asks what its function is. the answer is catalyst- but that doesn't make sense to me because wouldn't CuSO4 dissociate into copper II and sulfate, then the sulfate would act to inhibit the reaction by the common ion effect?

thanks!

 

[rcd]

another question...

rxn 1: 2I (-) + S2O8 (2-) --> I2 + 2SO4 (2-)

there is a table with rxn rates at various concentrations of the reactants and at different temperatures. when CuSO4 is added, the rxn time is cut in half, and the question asks what its function is. the answer is catalyst- but that doesn't make sense to me because wouldn't CuSO4 dissociate into copper II and sulfate, then the sulfate would act to inhibit the reaction by the common ion effect?

thanks!

You're mixing up kinetics with equilibrium.

 

[burtsicle]

So question asks:
Which of the following will be more soluble in 1.0 M HCl than in 1.0 M NaOH?

I choose the answer HI but the correct answer was PbOH2. The answer key claimed it was because of Le Chatlier's prinicple where PbOH2 would form water drawing the reaction to the right. BUT whatever happened to like dissolves like? i.e. HI and HCl are pretty similar??

Ideas?

 

[corbis11]

I think the acid is less soluble due to the common ion effect...

 

[BrokenGlass]

another question...

rxn 1: 2I (-) + S2O8 (2-) --> I2 + 2SO4 (2-)

there is a table with rxn rates at various concentrations of the reactants and at different temperatures. when CuSO4 is added, the rxn time is cut in half, and the question asks what its function is. the answer is catalyst- but that doesn't make sense to me because wouldn't CuSO4 dissociate into copper II and sulfate, then the sulfate would act to inhibit the reaction by the common ion effect?

thanks!

The key here is "cuts reaction time in half." So it has to be a catalyst.

 

[xlr8]

Hi

NaCl dissolves spontaneously in water. Based upon the following reaction:

NaCl (s) ---> Na+ (g) + Cl- (g) delta-H = 786 kJ/mol

The heat of hydration for NaCl must be:

A) negative magnitude less than 786.
B) negative with a magnitude greater than 786.
C) positive with a magnitude greater than 786.
D) Nothing can be determined without more information.


The way I looked at it is:
dH(soln) = dH(solute-solute) + dH(solvent-solvent) + dH(solute-solvent), where heat of hydration dH(hyd) = dH(solvent-solvent) + dH(solute-solvent).

So: dH(soln) = dH(solute) + dH(hyd) = 786 kj/mol

According to EK, dH(solute) is an endothermic step to break the solute-solute bonds, making it have a positive dH. This means that 786 = positive number + heat of hydration, thus the heat of hydration has to be positive and less than 786 according to that.

However, in this problem it says that the dH for the solute is indeterminable?

 

[BrokenGlass]

Hi

NaCl dissolves spontaneously in water. Based upon the following reaction:

NaCl (s) ---> Na+ (g) + Cl- (g) delta-H = 786 kJ/mol

The heat of hydration for NaCl must be:

A) negative magnitude less than 786.
B) negative with a magnitude greater than 786.
C) positive with a magnitude greater than 786.
D) Nothing can be determined without more information.


The way I looked at it is:
dH(soln) = dH(solute-solute) + dH(solvent-solvent) + dH(solute-solvent), where heat of hydration dH(hyd) = dH(solvent-solvent) + dH(solute-solvent).

So: dH(soln) = dH(solute) + dH(hyd) = 786 kj/mol

According to EK, dH(solute) is an endothermic step to break the solute-solute bonds, making it have a positive dH. This means that 786 = positive number + heat of hydration, thus the heat of hydration has to be positive and less than 786 according to that.

However, in this problem it says that the dH for the solute is indeterminable?

Rephrase your question. I have a hard time understanding it.

 

[BerkReviewTeach]

Hi

NaCl dissolves spontaneously in water. Based upon the following reaction:

NaCl (s) ---> Na+ (g) + Cl- (g) delta-H = 786 kJ/mol

The heat of hydration for NaCl must be:

A) negative magnitude less than 786.
B) negative with a magnitude greater than 786.
C) positive with a magnitude greater than 786.
D) Nothing can be determined without more information.


The way I looked at it is:
dH(soln) = dH(solute-solute) + dH(solvent-solvent) + dH(solute-solvent), where heat of hydration dH(hyd) = dH(solvent-solvent) + dH(solute-solvent).

So: dH(soln) = dH(solute) + dH(hyd) = 786 kj/mol

dH(solute) is an endothermic step to break the solute-solute bonds, making it have a positive dH. This means that 786 = positive number + heat of hydration, thus the heat of hydration has to be positive and less than 786 according to that.

However, in this problem it says that the dH for the solute is indeterminable?

Spontaneous describes the change in free energy, not enthalpy. The question never stated whether or not the overall process was endothermic or exothermic. As such, you have the following:

dH(rxn) = dH(dissociation) + dH(hydration)

which becomes: dH(rxn) = 786 + dH(hydration)

Two unknowns, so there is no way to solve the question based on the iformation they gave you.

Be careful not to interchange dG and dH.

 

[BerkReviewTeach]

So question asks:
Which of the following will be more soluble in 1.0 M HCl than in 1.0 M NaOH?

I choose the answer HI but the correct answer was PbOH2. The answer key claimed it was because of Le Chatlier's prinicple where PbOH2 would form water drawing the reaction to the right. BUT whatever happened to like dissolves like? i.e. HI and HCl are pretty similar??

Ideas?

This is a VERY common error people make. Like dissolves like refers to a solvent impacting solute. In both a 1.0 M HCl solution and a 1.0 M NaOH solution, the solvent is water, into which ions can readily dissociate. So, this question is not a like-dissolves-like question. This question is a case of complex equilibrium and/or common ion effect, depending on your perspective.

The H+ can protonate the hydroxides of Pb(OH)2, pushing the dissociation of Pb(OH)2 in the forward directions. This is the principle behind antacids. They are insoluble in water (allowing you to take them as a solid), but they dissolve in the low pH of gastric fluids. Acid consummes the base.

 

[xlr8]

Spontaneous describes the change in free energy, not enthalpy. The question never stated whether or not the overall process was endothermic or exothermic. As such, you have the following:

dH(rxn) = dH(dissociation) + dH(hydration)

which becomes: dH(rxn) = 786 + dH(hydration)

Two unknowns, so there is no way to solve the question based on the iformation they gave you.

Be careful not to interchange dG and dH.

the dH given, which is next to the reaction is not the dH of the reaction?

 

[BrokenGlass]

the dH given, which is next to the reaction is not the dH of the reaction?

It's a dissocation reaction, so why isn't dH given next to the reaction not the dH(dissociation)?

Hydration is the next step...

 

[xlr8]

It's a dissocation reaction, so why isn't dH given next to the reaction not the dH(dissociation)?

Hydration is the next step...

I see your point. So for every reaction in which a compound breaks down into ions in the aqueous phase I am to assume the dH given is that of dissociation, and not of the entire rxn?

 

[BrokenGlass]

I see your point. So for every reaction in which a compound breaks down into ions in the aqueous phase I am to assume the dH given is that of dissociation, and not of the entire rxn?

delta H written next to any reaction is referring to that reaction.

 

[BerkReviewTeach]

the dH given, which is next to the reaction is not the dH of the reaction?

It's the dH for the reaction listed, as Brokenglass pointed out.

That number is the sum of the conversion (sublimation and ionization) of sodium metal into a gaseous cation and the conversion of chlorine gas into a single gaseous chloride anion (half the BDE for a Cl2 bond and the electon affinity of a single chlorine atom).

This question is a spin-off of the Born-Haber cycle for salt formation. If you want a more detailed account, either look up Born-Haber in the index of your general chemistry text, or if you happen to have the BR general chemistry books, look at page 170 of Book II and the corresponding answer explanations.

 

[BirenJ]

I'm using EK's chem book to review gen chem. I need more practice with chem questions. What would you folks recommend? EK's 1001 questions? Anything else besides EK? Thanks.

 

[xlr8]

For some reason i thought that it was (aq) instead of (g) next to the Na+ and Cl-. If it was (aq), then would the dH represent that of dissociation and solvation?

 

[philios]

i was wondering how much of recognizing reaction orders we will have to know for the mcat. For reference I am looking at Examkrackers Gen. Chemistry, section 2-8, and wondering if we have to know those graphs on page 33. Thanks!

Also how much do we have to know on partial pressure equilibrium constants? these are pretty confusing for me, so I was wondering if they are on the MCAT at a minimum or a lot? Thanks for the help, I appreciate it.

 

[BrokenGlass]

i was wondering how much of recognizing reaction orders we will have to know for the mcat. For reference I am looking at Examkrackers Gen. Chemistry, section 2-8, and wondering if we have to know those graphs on page 33. Thanks!

Also how much do we have to know on partial pressure equilibrium constants? these are pretty confusing for me, so I was wondering if they are on the MCAT at a minimum or a lot? Thanks for the help, I appreciate it.

Which aspect of partial pressure equilibrium constant are you having trouble with? Please post a more specific question. It's just using partial pressures instead of concentrations in the equilibrium expression...

 

[scotties123]

easy question that i for some reason cant find the answer to. Is a pH indicator's range pKa+/- 1 or pH +/- 1?

 

[gridiron]

easy question that i for some reason cant find the answer to. Is a pH indicator's range pKa+/- 1 or pH +/- 1?

The range is within the (+/-)1 of the PKin (indicator) value.