General Chemistry Question Thread

[QofQuimica]

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Unacceptable topics:
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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.

 

[BrokenGlass]

Just say there is a polyprotic acid:

H2SO4

First Ka is the dissociation of H2SO4 -> H+ and HSO4-

Second HSO4 -> H+ + SO4-

And then it asked which quantities are equal...

They answered that H+ of the first rxn and H+ of the second rxn are equal! How can this be?

Are they assuming that since the 2nd reaction is so infavorable that basically the 1st rxn H+ will equal the 2nd rxn H+ because basically the 2nd rxn is creating negligible amounts of protons?

POLYPROTIC ACIDS

Polyprotic acids are acids with more than one acidic (removable) proton per molecule.
The protons are removed in a stepwise manner, one proton at a time.

Consider sulfuric acid, H2S04

It dissociates twice (diprotic). The conjugate base of the first dissociation
equilibrium becomes the acid in the second step.

(1) H2SO4(aq) + H2O(l) ---> H3O+(aq) + HSO4-(aq) Ka1 = very large
(2) HsO4-(aq) + H2O(l) <--> H3O+(aq) + SO42-(aq) Ka2 = 1.2 * 10^-2


Each dissociation step of a polyprotic acid has its own dissociation constant Ka.
The Ka always gets smaller with each ionization step. In general, Ka1 > Ka2 > Ka3,
etc. because as the negative charge on the acid increases, it becomes more difficult
to tear away the positively charged proton. Therefore, each acid formed from a stepwise
dissociation is weaker than the previous acid it came from.


H2SO4 is a strong acid (see below), HSO4- is a weak acid.


When we're trying to determine the pH of WEAK polyprotic acids, we just have to consider
the first reaction (i.e. treat them as weak monoprotic acids) even though H3O+ is produced
in subsequent reactions because:

(a) the ionization constants for the second and subsequent reactions are going to be
super-small anyway (Ka1 >> Ka2 > Ka3)
(b) since it is a weak polyprotic acid, not much of the second acid is produced from
the first reaction anyway


An exception is sulfuric acid, the only common polyprotic acid that is a strong acid in its
first deprotonation. H2S04 can have large dissociations from 2nd ionization under right conditions.



===================================================================================================

SULFURIC ACID IS A STRONG ACID


A strong acid is an acid that is (almost) completely ionized in aqueous solution.
A weak acid is an acid that only partially ionizes in aqeuous solution.
Acid strength depends on bond polarity, bond strength, and stability of conjugate base.


In addition to being a polyprotic acid, H2SO4 is an oxyacid. Oxyacids are acids with
the ionizable proton attached to oxygen. An oxoacid has the structure H-O-Y. In other
words, an acidic H atom is always attached to an O atom, which in turn is attached to
another atom Y.


Since each oxyacid has an O-H bond, we cannot distinguish their acid strength based on
bond polarity or bond strength. So lets consider conjugate base stability.
Relative acid strength of oxyacids depends on:

(a) Number of oxygen atoms around the central atom. For oxyacids that have the
same central atom Y, acid strength increases as the number of oxygen atoms
attached to Y increases. This is due to resonance stabilization of conjugate
base.
(b) Electronegativity of Y (the central atom). For oxyacids that have the same
number of OH groups and the same number of O atoms, acid strength increases
with increasing electronegativity of the central atom. This is due to electron
withdrawing inductive effect which results in stabilization of conjuage base.


OH
|
O==S==O
|
OH

Hopefully this provides some intuition why H2SO4 is a strong acid.

=====================================================================================================
PERCENT DISSOCIATION AND EFFECT OF ACID CONCENTRATION ON % DISSOCIATION


Percent ionization is another method to access acid strength. It relates equilibrium
H+ concentration to the initial HA concentration. The higher percent ionization,
the stronger the acid. Strong acids are 100% ionized. Only partial ionization occurs
with weak acids


% dissociation gives us an idea of the amount of weak acid that has dissociated.


% dissociation = ( [HA] dissociated / [HA] initial ) * 100 %


A strong acid dissociates completely (100%) regardless of initial concentration.
The % dissociation of weak acids varies with acid concentration. Although complete
dissociation is usually associated with "strong" acids, at low concentration, "weak"
acids behave like "strong" acids. Contributions of H+ from HSO4- can be substantial in
dilute solutions.


As the solution of a WEAK acid becomes more dilute, the percent dissociation increases.


Le Chaterlier's principle can be used to explain why % dissocation increases as the
concentration of a weak acid decreases. Let's look at the equilibrium of a weak acid:


HX(aq) + H2O <--> H3O+(aq) + X-(aq)

From Le Chatelier's Principle, adding water to the equilibrium would cause the equilibrium
to shift to the right. A shift to the right implies that more acid would be in dissociated
form, and thus the percent ionization increases accordingly.


Do not confuse a weak acid with a dilute acid. A weak acid has a small Ka, and a dilute acid has
a low concentration. It is possible to have a dilute, strong acid or a concentrated, weak acid.

======================================================================================================

SULFURIC ACID IS A UNIQUE POLYPROTIC ACID


Sulfuric acid is different from other polyprotic acids in 2 ways:

1) Ka1 = very large
2) Ka2 = 1.2 * 10^-2 (i.e. large for a weak acid, so HSO4- is a "strong" weak acid)


H2SO4(aq) + H2O(l) ---> H3O+(aq) + HSO4- (aq) Ka1 >> 0 (Ka1 is very large, so H2SO4 dissociates completely)
HsO4-(aq) + H2O(l) <==> H3O+(aq) + SO42- (aq) Ka2 = 1.2 * 10^-2


If [H2S04] > 1.0M (perhaps even 0.5M), the pH is solely determined from the first acidic proton, but if it
is more dilute, we need to account for the second dissociation contributing some H+ and changing pH.

Here are the details:

-In concentrated solutions ([H2S04] > 0.50M), the first ionization produces all the H3O+.
-In intermediate concentrations (0.0010 to 0.50 M), consider both ionizable H+ ions (do an ICE
box for 2nd ionization and solve a quadratic equation).
-In dilute solutions ([H2S04] < 0.0010 M), the second dissociation goes to completion (so assume
complete dissociation of 2H+ ions). I am guessing that this is the scenario your question was asking about.

====================================================================================================

EXAMPLES



Example 1: Evaluate the pH of 1.0 M H2SO4.



H2SO4 (aq) -> H+ (aq) + HSO4- (aq) (complete dissociation of the first proton)
Initial(M) 1.0 0 0
Change (M) -1.0 +1.0 +1.0
Equilibrium (M) 0 1.0 1.0

pH = -log (1.0) = 0 (after complete dissocation of first proton)



HSO4- (aq) <--> H+ (aq) + SO42- (aq)
Initial(M) 1.0 1.0 0
Change (M) -x +x +x
Equilibrium (M) 1.0 &#8211; x 1.0 + x x


Ka2 = ([H+][SO42-]) / [HSO4-]

1.2 * 10^-2 = ((1.0 + x) * x) / (1.0 &#8211; x) (neglect x because x is very small)

x = 0.012 M

[H+] = 1.0 M + 0.012 M = 1.012M
pH = -log [H+] = - log (1.012) = 0.005


Note that pH of 1.0M H+ is already 0.00. Second dissociation didn't change pH all that much.


% dissociation = (0.012 / 1.0)* 100 % = 1.2%




Example 2: Evaluate the pH of 0.0100 M H2SO4.


H2SO4 (aq) -> H+ (aq) + HSO4- (aq) (complete dissociation of the first proton, Ka1 is large, strong acid )
Initial 0.0100 M 0 M 0 M
Final 0 0.0100 M 0.0100 M

pH = -log (0.01) = 2 (after complete dissociation of first proton)

HSO4- (aq) <--> H+ (aq) + SO42- (aq) (Ka2 = 1.2 * 10 ^-2)
Initial 0.0100 M 0.0100 M 0 M
Final 0.0100 M - x 0.0100 M + x x

Ka = 0.012 = ([H+][SO4-2]) / [HSO4-]

1.2 * 10^-2 = ((0.0100 + x)*x) / (0.0100&#8211; x) can't neglect x, quadratic must be utilized


x^2 + 0.022 x &#8211; 0.00012 = 0
x = 0.004524 M
[H+] = 0.0100 M + 0.004524 M = 0.014524 M
pH = -log [H+] = - log [0.014524] = 1.840


% dissociation = ( 0.004524 / 0.0100 ) * 100% = 45.24%

conclusion: Contributions of H+ from HSO4- can be substantial in dilute solutions.

Sorry for the long post

 

[chem neub]

Le Chaterlier's principle according to teacher/ equilibrium is only dependent on temperature so could someone please explains shifts based on pressure heres a question
If the volume of the reaction container is increased shifts will be Right/ Left/ None


N2 +3H2 <--> 2NH3 (all gases)
PCl5 <--> PCl3 +Cl2 (all gases)

 

[gridiron]

Le Chaterlier's principle according to teacher/ equilibrium is only dependent on temperature so could someone please explains shifts based on pressure heres a question
If the volume of the reaction container is increased shifts will be Right/ Left/ None


N2 +3H2 <--> 2NH3 (all gases)
PCl5 <--> PCl3 +Cl2 (all gases)

Hey! Le Chatelier's principle is not only dependent on temperature. By definition, a system at equilibrium will want to stay in equilibrium and will try to neutralize changes to come back to equilibrium. Disturbances include changing the concentration of the species, changing the pressure and temperature. If the volume is increased, the pressure is decreased. That means the pressure of the system will increase as a result to come back to equilibrium. How will that happen? Pressure will increase on the side with more moles of gases. That means for the first equation, the equation will shift left. For the second reaction, this means it will shift right. What would happen if the volume were decreased? That means the pressure increases so the equation will shift to the side with the least amount of moles of gas in order to reduce the pressure. A good analogy to understanding Le Chatelier's principle is Newton's first law. Hope this helps and good luck .

 

[Onegin]

I have a somewhat basic question... what is the best way to memorize the various electron configurations for the transition metals? Thanks

 

[killinsound]

What is the effect of increasing the concentration of reactants in a voltaic cell?


My thought process is this:

When you are putting together different redox potentials

ie: A2+ + 2 e- -> A E= .2

B+ + e- -> B E= -.1

So the E in this case is .3. Even when you multiply the 2nd equation by two, the E does not change.

Therefore, I concluded that if you increase the concetrations of reactants, nothing will happen. However, the answer says that both the voltage and the spontaneity of the reaction will increase.

Can someone tell me why? What's the difference between waht the question is asking and how I am deducing?

Thanks

 

[amestramgram]

I have a somewhat basic question... what is the best way to memorize the various electron configurations for the transition metals? Thanks

if you mean Fe3+, Fe2+, Mn7+?

I just wrote them out over and over again. Nobody was kind enough to tell me a mnemonic for them *tear*

@ kilinsound:

E = E0 -(RT/nF) ln Q

reaction: aA + bB --> cC + dD

Q = (C^c)*(D^d) / ((A^a)*(B^b))

if you increase the denominator

lnQ becomes more negative

-(RT/nF) ln Q becomes more positive

and then you increased E

which makes it more spontaneous than before.

Also, you could have reason by le Châtelier's principle that if you add more reactants, the reaction is more likely to go towards products.

 

[killinsound]

if you mean Fe3+, Fe2+, Mn7+?

I just wrote them out over and over again. Nobody was kind enough to tell me a mnemonic for them *tear*

@ kilinsound:

E = E0 -(RT/nF) ln Q

reaction: aA + bB --> cC + dD

Q = (C^c)*(D^d) / ((A^a)*(B^b))

if you increase the denominator

lnQ becomes more negative

-(RT/nF) ln Q becomes more positive

and then you increased E

which makes it more spontaneous than before.

Also, you could have reason by le Châtelier's principle that if you add more reactants, the reaction is more likely to go towards products.

Thanks. However, I understand that, what I don't understand is what the difference between that, and adding the Electric Potential from reduction potentials. When you multiply those by two, you do not increase the electric potentia.

 

[amestramgram]

the E0 value shall not be multiplied by anything despite the fact you multiplied the half reaction by something.

because

it is an experimental observation that E0 values are intensive properties. Just like copper is yellowish brown no matter what size of copper you have in your hand - E0 is E0 because it is an intensive property of that half reaction.

here I found a really detailed proof 🙂

http://www.sparknotes.com/chemistry/electrochemistry/thermo/section1.html

 

[killinsound]

the E0 value shall not be multiplied by anything despite the fact you multiplied the half reaction by something.

because

it is an experimental observation that E0 values are intensive properties. Just like copper is yellowish brown no matter what size of copper you have in your hand - E0 is E0 because it is an intensive property of that half reaction.

here I found a really detailed proof 🙂

http://www.sparknotes.com/chemistry/electrochemistry/thermo/section1.html

Thanks alot amestramgram!! 🙂

 

[BrokenGlass]

What is the effect of increasing the concentration of reactants in a voltaic cell?


My thought process is this:

When you are putting together different redox potentials

ie: A2+ + 2 e- -> A E= .2

B+ + e- -> B E= -.1

So the E in this case is .3. Even when you multiply the 2nd equation by two, the E does not change.

Therefore, I concluded that if you increase the concetrations of reactants, nothing will happen. However, the answer says that both the voltage and the spontaneity of the reaction will increase.

Can someone tell me why? What's the difference between waht the question is asking and how I am deducing?

Thanks

Amestramgram already explained that the answer to this question follows from Nernst equation (E = E0 -(RT/nF) ln Q and delta G = -nFE.
Clearly, E depends on concentration of the reactants and the products, so changing concentration of reactants while keeping all else fixed
changes E.

I would like to add the following comments:

It is true that potential (i.e. voltage) is an intensive property, so it doesn't depend on the amount. The reason is that voltage is
defined as potential energy PER UNIT CHARGE. Simialry, density is an intensive property because density is mass PER UNIT VOLUME
and pressure is an intensive property because pressure can be thought of as the random translational kinetic energy of molecules
PER UNIT VOLUME). So when your definition includes "per unit of something", you have defined an intensive property.

When you multiply a chemical equation by some constant, you are not necessarily increasing the concentration of reactants.
Coefficients represent moles, not concentrations. The concentration (e.g. moles/L) would increase only if volume stays the same
(or goes down).

 

[cemented]

sorry

 

[Crestone]

How do you convert pH to concentration of H+ without using a calculator? For example, suppose you have a solution with a pH of 4.32, and you want to know what is the concentration of H+. I know the answer is 4.8 x 10^-5, but how would I figure that out, at least approximately, without a calculator? Thanks very much for your help.

 

[BrokenGlass]

How do you convert pH to concentration of H+ without using a calculator? For example, suppose you have a solution with a pH of 4.32, and you want to know what is the concentration of H+. I know the answer is 4.8 x 10^-5, but how would I figure that out, at least approximately, without a calculator? Thanks very much for your help.

pH = -log[H+]
If you were given that [H+]= 4.8 x 10^-5, then pH would be 5 - log 4.8 = 5 - 0.68 = 4.32.
(Since log 5 = 0.7, log of 4.8 is a little less than 0.7; 4.8 is closer to 1 than than 5 is,
and since log of 1 is 0, log of 4.8 is closer to 0 than log of 5 is.

BTW, it's good to memorize that log 2 = 0.3 and log 3 = 0.5; many other logs can be derived
just from knowing these 2 values. For instance, log 5 = log (10/2) = log 10 - log 2 = 1 - 0.3 = 0.7).

What if you are given that pH = 4.32 and you are asked to find [H+]?
[H+] = 10^-pH = 10^-4.32 = ?
You know that your answer has to be between 10^-4 and 10^-5. Often this "ranging" is good
enough on the MCAT because there is probably only one answer choice that falls in this range,
so you would not have to make fine distinctions between answer choices.

But in case you do, you can take [H+] in each answer choice (e.g. 4.8 x 10^-5) and
figure out the pH for it using the reasoning I outlined above. The answer choice that
gives you the pH closest to 4.32 is the correct answer, which in this case happens to
be 4.8 x 10^-5

Finally, you can resort to doing this:

pH = -log[H+] = 4.32 (this is given)
If [H+]= 10^-5, then pH = 5; If [H+]= 10^-4, then pH = 4,
so [H+] should be some positive number times 10^-5. Lets call this nubmer y.
[H+] = y * 10^-5, so pH = 5 - log y = 4.32; 4.32 is approximately equal to 4.30.
So you know log y = 0.7, so y = 5 (see above), so [H+] = 5 x 10^-5, which is approximately
equal to 4.8 x 10^-5

 

[Crestone]

Thanks so much, brokenglass! That helps a lot.

 

[minsoox1983]

This passage is about solubility of Pb.


[Passage ]

A series of chemical reactions was carried out to study the chemistry of lead.

Reaction 1

Initially, 15.0 mL of 0.300 M Pb(NO3)2(aq) was mixed with 15.0 mL of 0.300 M Na2SO4(aq). All the Pb(NO3)2 reacted to form Compound A, a white precipitate. Compound A was removed by filtration.

Reaction 2

Next, 15.0 mL of 0.300 M KI(aq) was added to Compound A. The mixture was agitated and some of Compound A dissolved. In addition, a yellow precipitate of PbI2(s) was formed.

Reaction 3

The PbI2(s) was separated and mixed with 15.0 mL of 0.300 M Na2CO3(aq). A white precipitate of PbCO3(s) formed. All of the PbI2(s) was converted into PbCO3(s).

Reaction 4

The PbCO3(s) was removed by filtration and a small sample gave off a gas when treated with dilute HCl.


Q1)
A soluble form of Pb2+ can be carefully added to a solution to sequentially precipitate and separate anions present in the solution. When Pb2+ is added, in what order will the following anions be precipitated?

A) SO42- then I-

**B) CO32- then I-
The reactions described in the passage show that lead(II) is successively precipitated as PbSO4, PbI2, and PbCO3. This sequence shows (assuming equal anion concentrations, as must be done here) that PbCO3 is less soluble than PbI2, and PbI2 is less soluble than PbSO4. The order in which the anions precipitate Pb2+ is: CO32- then I- then SO42-. When this sequence is applied to the question, answer choice B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer choice B is the best answer.

C) SO42- then CO32-

D) I- then CO32-
---------------------------------------------

My question is,
1) When something precipitates,, is that mean that chemical is not soluble in the solvent ?


2) I still don't understand how they extracted the idea of "SO42- > CO32- > I-" from the passage. All I see is
Experiment #1 - PbSO4 is formed as a compound A, and precipitates
Experiment #2 - PbSO4 is transformed(?) to PbI2, and precipitates
Experiment #3 - PbI2 is transformed to PbCO3 and precipitates **it says "All PbI2 was converted to PbCO3." Is that mean PbCO3 is preferred to form more than PbI2 because of their solubility ?

Experiment #4 - PbCO3 treated with dilute HCl. It will make PbCl2 + H2CO3 right? but, what's the point ? why do they do that?

3) I thought PbX (Pb with any halogen) is insoluble in water isn't it?

Thanks!

 

[Onegin]

I have a quick question regarding galvanic cells. I admittedly did not catch alot of the electrochemistry the first time around and am having some trouble getting the ideas down. What exactly is the purpose of the salt bridge in the galvanic cell. I understand why two solutions are needed, to prevent the electrons from having a simpler path with less potential. But I still am not quite sure what the salt bridge is used for.

Thanks!

 

[gridiron]

Hey! The purpose of the salt bridge in a galvanic cell is to provide an electrical connection between the reduction and oxidation vessels of the cells without allowing the vessel contents to mix. This allows for the electron transfer between the vessels so that charge doesn't buildup in the vessels--otherwise you would have the separation of charge which is essentially a capacitor.

 

[HelpingHand]

Hello,

I have a few questions (from Examkrackers Chemistry 1001):

1 (347): Why is W = 0 in a free adiabatic expansion of a gas? I understand that dE = -W because Q = 0, but I don't understand why W = 0.

2 (353): Why doesn't internal energy depend on pressure or volume for an ideal gas: Isn't dE = Q - PdV?

Thank you, and for future reference, should I browse through previous posts to avoid "repeats"? 😳

 

[gridiron]

Hello,

I have a few questions (from Examkrackers Chemistry 1001):

1 (347): Why is W = 0 in a free adiabatic expansion of a gas? I understand that dE = -W because Q = 0, but I don't understand why W = 0.

2 (353): Why doesn't internal energy depend on pressure or volume for an ideal gas: Isn't dE = Q - PdV?

Thank you, and for future reference, should I browse through previous posts to avoid "repeats"? 😳

Hey Helping Hand! During an adiabatic process, both the pressure and volume change so there is work. If you were considering the process of adiabatic expansion followed by adiabatic compression, then the work would be zero because you would be dealing with a cyclical integral. However, for free adiabatic expansion, the first law simplifies to:

dU= del W (not dW) because work is path dependent so you use an inexact differential. However, del W = PdV. To solve this, and I wouldn't think this would be on the MCAT, you need a relationship for dP and dV since the pressure and volume change. Also, I do believe the internal energy is dependent on pressure/volume for an ideal gas. The expression for work would be dependent on whether you are dealing with an isobaric, isochoric, isothermal or adiabatic process. Hope this helps!

 

[googlinggoogler]

Hi, why does the reaction: Zn + NaNO3 --> not proceed? Can somebody explain this to me? Thanks so much!

 

[BrokenGlass]

Hi, why does the reaction: Zn + NaNO3 --> not proceed? Can somebody explain this to me? Thanks so much!

Sodium nitrate reacts with zinc in sodium hydroxide solution to give ammonia and the tetrahydroxozinc(II) ion, Zn(OH)42– . It's a REDOX reaction.

 

[BrokenGlass]

Hi, why does the reaction: Zn + NaNO3 --> not proceed? Can somebody explain this to me? Thanks so much!

Standard reduction potential are defined with respect to hydrogen, whose standard reduction potential is arbitrarily assigned a value of zero.
Potential and voltage are the same thing. By definition, voltage is potential energy per unit charge and since potential energy has no absolute
value, we are free to choose any reference point we wish. In this case, standard reduction potential of hydrogen is chosen as a reference point and arbitrarily assigned the value of zero.

If you check the table of standard reduction potentials, you will seethe following values:

Standard reduction potential for Zn is -0.76 (the more negative this value, the less is the tendency to be reduced)
Standard reduction potential for NO3- is +0.93 (the more positive this value, the greater is the tendency to be reduced)

Again, tendency to be reduced (and tendency to be oxidized) are defined with respect to hydrogen.

Reduction is Gain of electrons, Oxidation is Loss of Electrons (LEO GER)

delta G = - n * F * E.

Delta G < 0 for a spontaneous reaction. Therefore, E > 0 for a spontaneous reaction. Since a redox reaction involves an oxidation half
and a reduction half, our goal is to flip one of the reduction half reactions so that we have an oxidation half and a reduction half reactions.
We would like the sum of standard reduction potential and standard oxidation potential to be positive since E > 0 for a spontaneous reaction.

If we flip the sign of -0.76, we get +0.76 for the value of standard oxidation potential of Zinc.

+0.76 + 0.93 > 0, so the resulting redox reaction is favorable (i.e. spotaneous). So it seems to me that this reaction would proceed.


It would help if you can post the entire problem statement and the answer key, if they are available. Maybe gridiron has a better explanation.

 

[minsoox1983]

[Question] A solution of H2SO4(aq) has a pH of 6.0. What is the H3O+(aq) concentration?

A) 1 x 106

B) 5 x 10-7

C) 2 x 10-6

*D) 1 x 10-6
The definition of pH is -log[H+]. If the pH is 6.0, then 6.0 = -log[H+], and [H+] = antilog (-6.0) = 10-6. Thus, answer choice D is the best answer.


I do understand pH = -log[H+]. But H2SO4 is a diprotic acid, in which it generates 2 H+ s. I don't see why not pH = -log 2*[H+] ? Isn't the two H+ result in pH of 6 for every 1 mole of H2SO4 ?

Thanks !
-minsoo.

 

[BrokenGlass]

[Question] A solution of H2SO4(aq) has a pH of 6.0. What is the H3O+(aq) concentration?

A) 1 x 106

B) 5 x 10-7

C) 2 x 10-6

*D) 1 x 10-6
The definition of pH is -log[H+]. If the pH is 6.0, then 6.0 = -log[H+], and [H+] = antilog (-6.0) = 10-6. Thus, answer choice D is the best answer.


I do understand pH = -log[H+]. But H2SO4 is a diprotic acid, in which it generates 2 H+ s. I don't see why not pH = -log 2*[H+] ? Isn't the two H+ result in pH of 6 for every 1 mole of H2SO4 ?

Thanks !
-minsoo.

We don't need to worry if it's a polyprotic acid or a monoprotic acid for this question. If the question was aksing about molarity of H2SO4(aq), then we would have to consider what kind
of acid it is (polyprotic or monoprotic). For this question the source of H+ ions is irrelevant. If you know the pH, you can calculate [H+]. If you know [H+] you can caclculate the pH
without knowing the source of H+ ions.

 

[googlinggoogler]

Hi BrokenGlass, thanks for the response! The actual question is:
"Which reaction does not proced?
a) Li + CuCl2 -->
B) Zn + NaNO3 -->
C) CaCO3 + HCl -->
D) K + H2O -->
E) H+ + OH- -->

The answer is B, and the solution given is that all the other ones react, so I was supposed to choose B from process of elimination. But after seeing the solution, it didn't help me very much because I didn't understand why B wouldn't proceed. Someone told me the reason may have something to do with the metal reactivity series, but I'm not quite sure.

 

[BrokenGlass]

Hi BrokenGlass, thanks for the response! The actual question is:
"Which reaction does not proced?
a) Li + CuCl2 -->
B) Zn + NaNO3 -->
C) CaCO3 + HCl -->
D) K + H2O -->
E) H+ + OH- -->

The answer is B, and the solution given is that all the other ones react, so I was supposed to choose B from process of elimination. But after seeing the solution, it didn't help me very much because I didn't understand why B wouldn't proceed. Someone told me the reason may have something to do with the metal reactivity series, but I'm not quite sure.

Does the problem statement specify activation energies for these reactions? A reaction could be thermodynamically favorable, yet for all intents and purposes it would not proceed if it's kinetically unfavorable. Can you post the entire answer key?

 

[BrokenGlass]

Hi BrokenGlass, thanks for the response! The actual question is:
"Which reaction does not proced?
a) Li + CuCl2 -->
B) Zn + NaNO3 -->
C) CaCO3 + HCl -->
D) K + H2O -->
E) H+ + OH- -->

The answer is B, and the solution given is that all the other ones react, so I was supposed to choose B from process of elimination. But after seeing the solution, it didn't help me very much because I didn't understand why B wouldn't proceed. Someone told me the reason may have something to do with the metal reactivity series, but I'm not quite sure.

The key to this problem is to NOT think of Na+ simply as counterion. We have to compare Na+ to Zn2+. NO3- is our anion, but which cation does it like better, Na+ or Zn2+?

The standard reduction potential for Zn2+ = -076, BUT the standard reduction potential for Na+ = -2.71. So the standard oxidation potential for Na = +2.76 and the standard oxidation potential for Zn = +0.76. So Na likes to be oxidized more than Zn likes to be oxidized. It makes sense if you think about it because Na needs to loose only 1 electron to achieve a noble gas configuration, so it will gladly give up an electron.

So making Na+ from Na is always more favorable than making Zn2+ from An. So NO3- will not let go of Na+ to bond with Zn2+. So the reaction doesn't proceed.

 

[unsung]

Okay guys, I just started EK chemistry, and ran into a bit of a conceptual problem (pg 7, for those who have the book):

1. Why does the effective nuclear charge (Zeff) increase moving from lithium to beryllium, BUT decrease moving from nitrogen to oxygen? (That is, beryllium feels a higher nuclear charge than lithium, while oxygen feels a lower nuclear charge than nitrogen).

To simplify, why is Zeff(Be) > Zeff(Li), while Zeff(O) < Zeff(N)?

I understand that going from lithium (1s2 2s1) to beryllium (1s2 2s2), the addition of an electron into the same subshell (2s) counteracts the blocking effect from the first subshell (1s). So that makes sense that the outer electron in lithium feels less of a pull toward the nucleus than the outer electron in beryllium.

BUT, by that same logic, shouldn't oxygen (1s2 2s2 2px2 2py1 2pz1) also have a greater Zeff than nitrogen (1s2 2s2 2px1 2py1 2pz1)?

That is, an electron in the outermost 2p orbital is "blocked" or "shielded" from some of the nuclear charge by the inner s orbitals (1s2 and 2s2). The difference between oxygen and nitrogen being that in oxygen, there's an additional electron in the outermost orbital (2p), while both feel the same shielding from inner orbitals (1s & 2s)...

So following the trend observed with Li & Be, shouldn't Zeff(O) > Zeff (N), just like Zeff(Be) > Zeff(Li)?

2. How does "shielding" ACTUALLY work in multielectron atoms? For example, in helium, does the first electron a) physically shield/block/absorb some of the nuclear charge of the proton from the electron (i.e. the second electron actually feels a single reduced charge from the nucleus),

OR, b) exert a force in the opposite direction of the nuclear force, thus sort of counteracting the force of the first electron, resulting in overall, a reduced force felt ? (i.e. the second electron feels BOTH the full nuclear charge as well as an opposing force from the first electron)

I know the result is the same no matter which mechanism (a or b) is actually correct. But it still bothers me that I don't quite have a grasp of the mechanism by which this effect works.

I hope my questions made sense... and that someone has an explanation! Thanks in advance!

 

[minsoox1983]

Q. When a strip of Cu is placed into H2O(l), no change is observed. However, when a strip of Cu is placed into a solution of HNO3(aq), a gas evolves. What is the most likely identity of the gas?

A) NO(g)
Nitrogen monoxide, NO, is one gas evolved when copper metal is placed in a nitric acid (HNO3) solution. This reaction is an oxidation-reduction reaction, where copper is oxidized (loses electrons) and nitrogen is reduced (gains electrons). Nitric acid is an oxidizing acid, and nitrate salts can be oxidizing agents as well, under the proper conditions. Answer B can be eliminated because there is no stated source for the carbon in CO2. Answer D can be eliminated since the formation of ozone (oxidation state for oxygen = 0) from HNO3 or H2O (oxidation state for oxygen = -2 in both compounds) would require that copper gain electrons, forming a negative oxidation state, which is quite unusual for metals. The formation of H2 (answer C) is plausible, but is not observed. The much more electronegative nitrogen undergoes reduction more readily than does hydrogen. Thus, answer choice A is the best answer.

B) CO2(g)

C) H2(g)

D) O3(g)

----------------------------------------------------------------------------
what i can see is that the overall reaction.

Cu + 2(HNO3) -> Cu(HNO3)2

so Cu (0) --> Cu (+2) oxidized.
and something reduced.. but i couldn't find it. so i assumed 2H (+1) --> H2 (0) . Why is it wrong? and where is the NO (g) coming from ? NO3 -> NO + O2 ?

Thanks for your help!
😍

 

[dochoov]

.Consider the following gas phase reaction:

H2 (g) + Br2 (g) <---> 2HBr (g) (Edited here--my chemdraw equation wouldn't copy)

The concentrations of H2, Br2, and HBr are 0.05M, 0.03M, and 500.0M, respectively. The equilibrium constant for this reaction at 400 C is 2.5E3. Is this system at equilibrium?

A. Yes, the system is at equilibrium.
B. No, the reaction must shift to the right in order to reach equilibrium.
C. No, the reaction must shift to the left in order to reach equilibrium.
D. It cannot be determined if this system is at equilibrium without more information.

Kaplan says C. Here's my confusion: I think the answer should be D because we don't know what temperature those reactant/product concentrations are at. Can someone else please provide some insight here.
.

 

[swifteagle43]

.Consider the following gas phase reaction:

H2 (g) + Br2 (g) <---> 2HBr (g) (Edited here--my chemdraw equation wouldn't copy)

The concentrations of H2, Br2, and HBr are 0.05M, 0.03M, and 500.0M, respectively. The equilibrium constant for this reaction at 400 C is 2.5E3. Is this system at equilibrium?

A. Yes, the system is at equilibrium.
B. No, the reaction must shift to the right in order to reach equilibrium.
C. No, the reaction must shift to the left in order to reach equilibrium.
D. It cannot be determined if this system is at equilibrium without more information.

Kaplan says C. Here's my confusion: I think the answer should be D because we don't know what temperature those reactant/product concentrations are at. Can someone else please provide some insight here.
.

They are at 400 * Celcius.....

 

[BrokenGlass]

.Consider the following gas phase reaction:.

.H2 (g) + Br2 (g) <---> 2HBr (g) (Edited here--my chemdraw equation wouldn't copy).

.The concentrations of H2, Br2, and HBr are 0.05M, 0.03M, and 500.0M, respectively. The equilibrium constant for this reaction at 400 C is 2.5E3. Is this system at equilibrium?.

.A. Yes, the system is at equilibrium..
.B. No, the reaction must shift to the right in order to reach equilibrium..
.C. No, the reaction must shift to the left in order to reach equilibrium..
.D. It cannot be determined if this system is at equilibrium without more information..

.Kaplan says C. Here's my confusion: I think the answer should be D because we don't know what temperature those reactant/product concentrations are at. Can someone else please provide some insight here..

Usually, you would do this type of question by comparing Q (the reaction quotent) with K (the equilibrium constant).

If Q < K, the reaction will shift right.
If Q > K, the reaction will shift left.
If Q = K, the system is already at equilibrium.

Q = 500^2/(0.05 * 0.03) > K = 2.5 * 10^-3

If Q > K, the reaction will shift to the left in order to reach equilibrium. Does the question mention the temperature at which the concentrations are given? It could just be a poorly worded question.

 

[dochoov]

Usually, you would do this type of question by comparing Q (the reaction quotent) with K (the equilibrium constant).

If Q < K, the reaction will shift right.
If Q > K, the reaction will shift left.
If Q = K, the system is already at equilibrium.

Q = 500^2/(0.05 * 0.03) > K = 2.5 * 10^-3

If Q > K, the reaction will shift to the left in order to reach equilibrium. Does the question mention the temperature at which the concentrations are given? It could just be a poorly worded question.

No, the question does not mention the tempurature at which the concentrations are given. The question that I posted is presented in its entirety. Their questions sometimes include little details like the one I thought I had found, to make sure that you are paying close attention to everything and all, but I agree with you in that this is probobly just a poorly worded question.

 

[swifteagle43]

No, the question does not mention the tempurature at which the concentrations are given. The question that I posted is presented in its entirety. Their questions sometimes include little details like the one I thought I had found, to make sure that you are paying close attention to everything and all, but I agree with you in that this is probobly just a poorly worded question.

Dude, the problem states 400 degrees celcius. You said "...reaction at 400 C is 2.5E3"

 

[Shrike]

Dude, the problem states 400 degrees celcius. You said "...reaction at 400 C is 2.5E3"

Dude, if it's really as worded the problem states what the K is at 400C, but does not state that that's temperature of this reaction right now. Yes, that's the implication, but you know quite well that implications don't cut it on the MCAT (except when they do).

Badly worded problem.

 

[dochoov]

Dude, the problem states 400 degrees celcius. You said "...reaction at 400 C is 2.5E3"

They say that K at 400C is 2.5E3. They do not specify what temperature the given concentrations are at. Obviously after seeing the answer I'm supposed to assume the concentrations were at 400C also. But given the way the question was worded, and the fact that Kaplan does often include tiny details that would change the answer if they are considered, I had reason to believe otherwise.

Edit: Sorry to repeat what Shrike said, I didn't read that post before posting.

 

[unsung]

I'm a little bit confused about how to determine whether a value given in a typical periodic table refers to, first of all, the mass number OR the molar mass, and secondly, whether the value is given in amu OR g/mol.

For example, for carbon, the number on the bottom (12.0) could refer to the mass number (# of protons + # of neutrons), in which case the value of 12.0 would not have any units. The number 12 could also refer to molar mass (i.e. atomic weight), in which case 12.0 would be in units of amu, and 12.0 amu would refer to the mass of ONE ATOM of carbon. Alternatively, the number 12 could also be understood as the number of grams of the element in ONE MOLE (i.e. 12 g of C per mole of C).

These are clearly quite different concepts, and it's strange to me that depending on the needs of the problem, sometimes I will use 12.0 in units of g/mol, and other times, treat it as quite a different concept (12 amu = mass of 1 atom).

Can anyone shed some light?

Edited: that should be NEUTRONS, of course! Pardon.. lol

 

[MSTPbound]

I'm a little bit confused about how to determine whether a value given in a typical periodic table refers to, first of all, the mass number OR the molar mass, and secondly, whether the value is given in amu OR g/mol.

For example, for carbon, the number on the bottom (12.0) could refer to the mass number (# of protons + # of electrons), in which case the value of 12.0 would not have any units. The number 12 could also refer to molar mass (i.e. atomic weight), in which case 12.0 would be in units of amu, and 12.0 amu would refer to the mass of ONE ATOM of carbon. Alternatively, the number 12 could also be understood as the number of grams of the element in ONE MOLE (i.e. 12 g of C per mole of C).

Well, you picked the right element because C-12 is the standard that all of the other elements are based off of. First thing, you probably know this, but the atomic mass printed on the periodic table is an average of all naturally occurring isotopes of an element. Just to be sure we are starting on the same page.

Now, with respect to carbon, 12 amu does NOT refer to the sum of protons and electrons... it refers to the sum of protons and NEUTRONS. So you can change the atomic mass to get the isotopic form of the same element by changing the quantity of NEUTRONS only... if you change the proton qty, you get a different element, because atomic number (Z) is given by the quantity of protons. Electrons are not in this discussion... we talk about electrons when we talk reactivity/rxns.

If a question asks you about the MOLAR MASS of a substance, they want to know how many grams of that substance are in one mole of it. How many grams carbon are in one mole of Carbon-12? 12 grams. How many grams are there in one mole of carbon-13? 13 grams. What is the molar mass of oxygen gas?

Well, oxygen is a diatomic molecule, so one mole of oxygen gas has (2x16) grams per mole = 32 g/mol.

Why does the amu correspond with grams per mol? Because a long time ago, it was decided that by convention and for convenience, amu would be assigned based on the mass of one twelfth of the mass of an unbound atom of the carbon-12 nucleus, at rest and in its ground state. In other words... that's just how it is, so accept it. 🙂

Just like we accept that a dozen is 12, whether we're talking about eggs or cement trucks.

These are clearly quite different concepts, and it's strange to me that depending on the needs of the problem, sometimes I will use 12.0 in units of g/mol, and other times, treat it as quite a different concept (12 amu = mass of 1 atom)...

Yes, it might seem strange, and if you REALLY wanted to know why, you can do some research on Dalton and Avogadro, but it won't help you on the MCAT. What you need to know is how to find the info you need to answer the question correctly.

They want amu? go to the periodic table.
They want molar mass? make SURE you know what species they're talking about (e.g. nitrogen gas as opposed to a nitrogen atom), go to the periodic table, rest ASSURED that since the problem didn't asked for any weird isotopes, that the MASS per MOLE is equal to the AMU per ATOM, and do your stoichiometry to answer the question.

The periodic table is a critical tool for you throughout the MCAT (and the rest of your life in chemistry)... that's why they let you (encourage you) to use it... it'll tell you molar mass, amu, reaction trends, size trends, electronegativity... etc. etc. etc. Ok sorry, I'm getting chem-happy now. Hope this helps!

-MSTPbound

 

[unsung]

Well, you picked the right element because C-12 is the standard that all of the other elements are based off of. First thing, you probably know this, but the atomic mass printed on the periodic table is an average of all naturally occurring isotopes of an element. Just to be sure we are starting on the same page.

Now, with respect to carbon, 12 amu does NOT refer to the sum of protons and electrons... it refers to the sum of protons and NEUTRONS. So you can change the atomic mass to get the isotopic form of the same element by changing the quantity of NEUTRONS only... if you change the proton qty, you get a different element, because atomic number (Z) is given by the quantity of protons. Electrons are not in this discussion... we talk about electrons when we talk reactivity/rxns.

If a question asks you about the MOLAR MASS of a substance, they want to know how many grams of that substance are in one mole of it. How many grams carbon are in one mole of Carbon-12? 12 grams. How many grams are there in one mole of carbon-13? 13 grams. What is the molar mass of oxygen gas?

Well, oxygen is a diatomic molecule, so one mole of oxygen gas has (2x16) grams per mole = 32 g/mol.

Why does the amu correspond with grams per mol? Because a long time ago, it was decided that by convention and for convenience, amu would be assigned based on the mass of one twelfth of the mass of an unbound atom of the carbon-12 nucleus, at rest and in its ground state. In other words... that's just how it is, so accept it. 🙂

Just like we accept that a dozen is 12, whether we're talking about eggs or cement trucks.



Yes, it might seem strange, and if you REALLY wanted to know why, you can do some research on Dalton and Avogadro, but it won't help you on the MCAT. What you need to know is how to find the info you need to answer the question correctly.

They want amu? go to the periodic table.
They want molar mass? make SURE you know what species they're talking about (e.g. nitrogen gas as opposed to a nitrogen atom), go to the periodic table, rest ASSURED that since the problem didn't asked for any weird isotopes, that the MASS per MOLE is equal to the AMU per ATOM, and do your stoichiometry to answer the question.

The periodic table is a critical tool for you throughout the MCAT (and the rest of your life in chemistry)... that's why they let you (encourage you) to use it... it'll tell you molar mass, amu, reaction trends, size trends, electronegativity... etc. etc. etc. Ok sorry, I'm getting chem-happy now. Hope this helps!

-MSTPbound

So it's just a matter of convention and how amu is defined. Phew. Okay... guess I just wanted some reassurance that it really DOES make sense that mass per mole is equivalent amu per atom... 😉

Now I can move forward & have peace of mind while doing my stoichiometry... Thanks!

 

[BrokenGlass]

So it's just a matter of convention and how amu is defined. Phew. Okay... guess I just wanted some reassurance that it really DOES make sense that mass per mole is equivalent amu per atom... 😉

Now I can move forward & have peace of mind while doing my stoichiometry... Thanks!

One amu is defined to be one twelfth of the mass one carbon-12 atom, where 12 denotes mass number.
So you can think of amu as mass of a proton or a neutron.

One mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 grams of carbon-12.

Here is a superficial explanation: both amu and mole are defined using carbon-12, which is why you can read atomic mass as either grams/mole or amu.

 

[5moreminutes]

Can someone explain me this statement, and what makes it true?

"There is about one hydrated H+ for every billion molecules in the glass."


Thanks
-Ria

 

[googlinggoogler]

Q: Which of the following will ocur if ammonia were added to a saturated solution of Na2SO4?
A: Ksp stays the same but more Na2SO4 dissolves.

I'm confused by this answer and the solution given. The solution says that when Na2SO4 dissolves, the solution will be acidic since sulfuric acid is formed. Therefore, the ammonia, which is a base, will react with the protons in solution, therefore more Na2SO4 can be dissolved.

What I don't understand is, I believe SO4^-2 will react with water to form H2SO4 + 2OH-. Therefore, although sulfuric acid is formed, I believe overall the solution is basic since 2 hydroxides are formed for every sulfuric acid that is formed. Since ammonia is also a base, I don't understand how that will allow more SO4^-2 to dissolve. In my opinion, adding a base to a basic solution will not help more conjugate base (SO4^-2) dissolve.

 

[MSTPbound]

Q: Which of the following will ocur if ammonia were added to a saturated solution of Na2SO4?
A: Ksp stays the same but more Na2SO4 dissolves.

I'm confused by this answer and the solution given. The solution says that when Na2SO4 dissolves, the solution will be acidic since sulfuric acid is formed. Therefore, the ammonia, which is a base, will react with the protons in solution, therefore more Na2SO4 can be dissolved.

What I don't understand is, I believe SO4^-2 will react with water to form H2SO4 + 2OH-. Therefore, although sulfuric acid is formed, I believe overall the solution is basic since 2 hydroxides are formed for every sulfuric acid that is formed. Since ammonia is also a base, I don't understand how that will allow more SO4^-2 to dissolve. In my opinion, adding a base to a basic solution will not help more conjugate base (SO4^-2) dissolve.

Yeah, I'm confused too. I hope someone answers this one. I can say this much though: it's not likely that H2SO4 and OH^- would easily coexist in solution; they would undergo rapid neutralization.

I also don't see how solvation of Na2SO4 creates an acidic solution. If anything, I would expect it to create a weakly basic solution:

SO42- + H2O <---> HSO4- + OH-

and I would still expect this equilibrium to lie left since it is more neutral on that side.

Now, suppose you add NH3; I suspect that if the above equilibrium lies left, that NH3 would tend to be deprotonated by SO42-. This would create more of the HSO4- species to the right which, I suppose, would create more "room" in the solution for sulfate ion solvation. The Ksp would stay the same because the solution would still be limited to solvation of a given amount of SO4- ion, but as NH3 consumes SO4- ion, more can be added (in otherwords, adding NH3 brings the Qsp down below the Ksp, so more SO4- would have to be added for Qsp to = Ksp). So I agree with the answer, but I don't know if I agree with the reasoning.

Hopefully more knowledgable responses will follow.

 

[BrokenGlass]

Can someone explain me this statement, and what makes it true?

"There is about one hydrated H+ for every billion molecules in the glass."


Thanks
-Ria

A hydrated proton is a hydronium ion, i.e. H3O+. This statment is probably implying that pH is very high because [H+] is very low.

 

[BrokenGlass]

Q: Which of the following will ocur if ammonia were added to a saturated solution of Na2SO4?
A: Ksp stays the same but more Na2SO4 dissolves.

I'm confused by this answer and the solution given. The solution says that when Na2SO4 dissolves, the solution will be acidic since sulfuric acid is formed. Therefore, the ammonia, which is a base, will react with the protons in solution, therefore more Na2SO4 can be dissolved.

What I don't understand is, I believe SO4^-2 will react with water to form H2SO4 + 2OH-. Therefore, although sulfuric acid is formed, I believe overall the solution is basic since 2 hydroxides are formed for every sulfuric acid that is formed. Since ammonia is also a base, I don't understand how that will allow more SO4^-2 to dissolve. In my opinion, adding a base to a basic solution will not help more conjugate base (SO4^-2) dissolve.

Message removed

 

[Shrike]

A hydrated proton is a hydronium ion, i.e. H3O+. This statment is probably implying that pH is very high because [H+] is very low.

pH isn't all that high: at pH = 7 there's 10^(-7) moles of hydronium per liter, but there are about 55 moles of water per liter, so that's about 1 hydronium per 550 million molecules. At one per billion, that'd be a pH of around 7.3.

 

[googlinggoogler]

Thank you Brokenglass for taking the time to answer my question. 👍

I have another question (hopefully more straight forward) on rate laws. I know that if experimental data is provided, you can find the rate law that way. But is it also true that if the slow step of a reaction mechanism is provided, that the rate law will have exponents equaling the coefficients?

For example:

xA + yB --> zC (slow step)

Is the rate law= k [A]^x ^y ?? (assuming that A and B are not intermediates)

 

[BrokenGlass]

Thank you Brokenglass for taking the time to answer my question. 👍

I have another question (hopefully more straight forward) on rate laws. I know that if experimental data is provided, you can find the rate law that way. But is it also true that if the slow step of a reaction mechanism is provided, that the rate law will have exponents equaling the coefficients?

For example:

xA + yB --> zC (slow step)

Is the rate law= k [A]^x ^y ?? (assuming that A and B are not intermediates)

If the rds is the first step, then that's correct (assuming the reaction is an elementary reaction). It's slightly more complicated when rds is not the first step.

 

[googlinggoogler]

Just so I'm sure I understand, when you say it gets more complicated, you mean, I would have to write the intermediates in terms of the initial reactants? So even if it wasn't the first elementary step, it still holds true (rate=k [A]^x ^y), I would just have to write the [A] and in terms of the initial reactants. ??

 

[dochoov]

This question comes from a passage. The following reaction has reached equilibrium at 1200K:

CO2 (g) + C (s) <--> 2CO (g)

The question:

When the system stabilized at 1200K, a sample of helium was injected into the furnace. What should happen to the amount of carbon dioxide in the system?

A. It should increase.
B. It should decrease.
C. It should be completely converted to carbon monoxide
D. It will remain the same.

Kaplan answer: D. Explanation: You should know that helium, a noble gas, is very unreactive and would almost certainly not react with any of the species in the furnace. Because the helium does not react with any of the species that participate in the equilibrium, the equilibrium is unaffected by the addition of helium. Even though it increases the total pressure inside the system, the partial pressures of the reacting gases are unchanged (Dalton's law) and therefore they keep on behaving as if the helium weren't present. The correct selection is therefore choice D.

The bold is what is really confusing me. I thought the answer should be A because an increase in pressure would cause the reaction to shift to the left (Le Chat.'s principle). And this whole thing about the partial pressures being the same... well, I'm confused.